Undecidable Translational Tilings with Only Two Tiles, Or One Nonabelian Tile

UNDECIDABLE TRANSLATIONAL TILINGS WITH ONLY TWO TILES, OR ONE NONABELIAN TILE

RACHEL GREENFELD AND TERENCE TAO

2 Abstract. We construct an example of a group G = Z × G0 for a finite abelian group G0, a subset E of G0, and two finite subsets F1,F2 of G, such that it is undecidable in ZFC whether Z2 × E can be tiled by translations of F1,F2. In particular, this implies that this tiling problem is aperiodic, in the sense that (in the standard universe of ZFC) there exist translational tilings of E by the tiles F1,F2, but no periodic tilings. Previously, such aperiodic or undecidable translational tilings were only constructed for sets of eleven or more tiles (mostly in Z2). A similar construction also applies for G = Zd for sufficiently large d. If one allows the group G0 to be non-abelian, a variant of the construction produces an undecidable translational tiling with only one tile F . The argument proceeds by first observing that a single tiling equation is able to encode an arbitrary system of tiling equations, which in turn can encode an arbitrary system of certain functional equations once one has two or more tiles. In particular, one can use two tiles to encode tiling problems for an arbitrary number of tiles.

1. Introduction

1.1. A note on set-theoretic foundations. In this paper we will be dis- cussing questions of decidability in the Zermelo–Frankel–Choice (ZFC) axiom system of set theory. As such, we will sometimes have to make distinctions between the standard universe1 U of ZFC, in which for instance the natural numbers N = NU are the standard natural numbers {0, 1, 2,... }, the integers Z = ZU are the standard integers {0, ±1, ±2}, and so forth, and also nonstan- ∗ dard universes U of ZFC, in which the model NU∗ of the natural numbers may possibly admit some nonstandard elements not contained in the standard natural numbers NU, and similarly for the model ZU∗ of the integers in this universe. However, every standard natural number n = nU ∈ N will have a well-defined counterpart nU∗ ∈ NU∗ in such universes, which by abuse of notation we shall arXiv:2108.07902v1 [math.CO] 17 Aug 2021 usually identify with n; similarly for standard integers. If S is a first-order sentence in ZFC, we say that S is (logically) undecidable if it cannot be proven within the axiom system of ZFC. By the Gödelcompleteness theorem, this is equivalent to S being true in some universes of ZFC while being false in others. For instance, if S is a undecidable sentence that involves the group Zd for some standard natural number d, it could be that S holds for the d d standard model Z = ZU of this group, but fails for some non-standard model d ZU∗ of the group.

1Here we of course make the metamathematical assumption that the standard universe exists, so that in particular ZFC is consistent. Of course, by the G¨odelincompleteness theorem, this latter claim, if true, cannot be proven within ZFC itself. 1 2 RACHEL GREENFELD AND TERENCE TAO

Remark 1.1. In the literature the closely related concept of algorithmic undecidability from computability theory is often used. By a problem S(x), x ∈ X we mean a sentence S(x) involving a parameter x in some range X that can be encoded as a binary string. Such a problem is algorithmically undecidable if there is no Turing machine T which, when given x ∈ X (encoded as a binary string) as input, computes the truth value of S(x) (in the standard universe) in ﬁnite time. One relation between the two concepts is that if the problem S(x), x ∈ X is algorithmically undecidable then there must be at least one instance S(x0) of this problem with x0 ∈ X that is logically undecidable, since otherwise one could evaluate the truth value of a sentence S(x) for any x ∈ X by running an algorithm to search for proofs or disproofs of S(x). Our main results on logical undecidability can also be modiﬁed to give (slightly stronger) algorithmic undecidability results; see Remark 1.12 below. However, we have chosen to use the language of logical undecidability here rather than algorithmic undecidability, as the former concept can be meaningfully applied to individual tiling equations, rather than a tiling problem involving one or more parameters x.

In order to describe various mathematical assertions as first-order sentences in ZFC, it will be necessary to have the various parameters of these assertions presented in a suitably “explicit” or “definable” fashion. In this paper, this will be a particular issue with regards to finitely generated abelian groups G = (G, +). Define an explicit finitely generated abelian group to be a group of the form d Z × ZN1 × · · · × ZNm (1.1) for some (standard) natural numbers d, m and (standard) positive integers N1,...,Nm, where we use ZN := Z/NZ to denote the standard cyclic group of 2 20 order N. For instance, Z × Z21 is an explicit finitely generated abelian group. We define the notion of a explicit finite abelian group similarly by omitting the Zd factor. From the classification of finitely generated abelian groups, we know that (in the standard universe U of ZFC) every finitely generated abelian group is (abstractly) isomorphic to an explicit finitely generated abelian group, but the advantage of working with explicit finitely generated abelian groups is that such groups G are definable in ZFC, and in particular have counterparts GU∗ in all universes U∗ of ZFC, not just the standard universe U.

1.2. Tilings by a single tile. If G is an abelian group and A, F are subsets of G, we deﬁne the set A ⊕ F to be the set of all sums a + f with a ∈ A, f ∈ F if all these sums are distinct, and leave A ⊕ F undeﬁned if the sums are not distinct. Note that from our conventions we have A ⊕ F = ∅ whenever one of A, F is empty. Given two sets F,E in G, we let Tile(F ; E) denote the tiling equation2 X ⊕ F = E (1.2) where we view the tile F and the set E to be tiled as given data and the indeterminate variable X denotes an unknown subset of G. We will be interested

2In this paper we use tiling to refer exclusively to translational tilings, thus we do not permit rotations or reﬂections of the tile F . Also, we adopt the convention that an equation such as (1.2) is automatically false if one or more of the terms in that equation is undeﬁned. UNDECIDABLE TILINGS WITH ONLY TWO TILES 3 in the question of whether this tiling equation Tile(F ; E) admits solutions X = A, and more generally what the space

Tile(F ; E)U := {A ⊂ G : A ⊕ F = E} of solutions to Tile(F ; E) looks like. Later on we will generalize this situation by considering systems of tiling equations rather than just a single tiling equation, and also allow for multiple tiles F1,...,FJ rather than a single tile F . We will focus on tiling equations in which G is a finitely generated abelian group, F is a finite subset of G, and E is a subset of G which is periodic, by which we mean that E is a finite union of cosets of some finite index subgroup of G. In order to be able to talk about the decidability of such tiling problems we will need to restrict further by requiring that G is an explicit finitely generated abelian group in the sense (1.1) discussed previously. The finite set F can then be described explicitly in terms of a finite number of standard integers; for 2 instance, if F is a finite subset of Z × ZN , then one can write it as

F = {(a1, b1, c1 mod N),..., (ak, bk, ck mod N)} for some standard natural number k and some standard integers a1 , ... , ak, b1, ... , bk, c1, ... , ck. Thus F is now a definable set in ZFC and has counterparts ∗ FU∗ in every universe U of ZFC. Similarly, a periodic subset E of an explicit d finitely generated abelian group Z × ZN1 × · · · × ZNm can be written as d E = S ⊕ ((rZ ) × ZN1 × · · · × ZNm ) for some standard natural number r and some finite subset S of G; thus E is ∗ also definable and has counterparts EU∗ in every universe U of ZFC. One can now consider the solution space

Tile(F ; E)U∗ := {A ⊂ GU∗ : A ⊕ FU∗ = EU∗ } to Tile(F ; E) in any universe U∗ of ZFC. We now consider the following two properties of the tiling equation Tile(F ; E). Definition 1.2 (Undecidability and aperiodicity). Let G be an (explicit) finitely generated abelian group, F a finite subset of G, and E a periodic subset of G. (i) We say that the tiling equation Tile(F ; E) is undecidable if the assertion that there exists a solution A ⊂ G to Tile(F ; E), when phrased as a first-order sentence in ZFC, is not provable within the axiom system of ZFC. By the Gödelcompleteness theorem, this equivalent to the assertion ∗ that Tile(F ; E)U∗ is non-empty for some universes U of ZFC, but empty for some other universes. We say that the tiling equation Tile(F ; E) is decidable if it is not undecidable. (ii) We say that the tiling equation Tile(F ; E) is aperiodic if, when working within the standard universe U of ZFC, the equation Tile(F ; E) admits a solution A ⊂ G, but that none of these solutions are periodic. That is to say, Tile(F ; E)U is non-empty but contains no periodic sets. Example 1.3. Let G be the explicit finitely generated abelian group G := Z2, let F := {0, 1}2, and let E := Z2. The tiling equation Tile(F ; E) has multiple solutions in the standard universe U of ZFC; for instance, given any (standard) function a: Z → {0, 1}, the set A := {(n, a(n) + m): n, m ∈ 2Z} 4 RACHEL GREENFELD AND TERENCE TAO solves the tiling equation Tile(F ; E) and is thus an element of Tile(F ; E)U. Most of these solutions will not be periodic, but for instance if one selects the function a ≡ 0 (so that A = (2Z)2) then one obtains a periodic tiling. This latter tiling is definable and thus has a counterpart in every universe U∗ of ZFC, and we conclude that in this case the tiling equation Tile(F ; E) is decidable and not aperiodic. Remark 1.4. The notion of aperiodicity of a tiling equation Tile(F ; E) is only interesting when E is itself periodic, since if A ⊕ F = E and A is periodic then E must necessarily be periodic also.

A well-known argument of Wang (see [B66, R71]) shows that if a tiling equation Tile(F ; E) is not aperiodic, then it is decidable; contrapositively, if a tiling equation is undecidable, then it must also be aperiodic. From this we see that any undecidable tiling equation must admit (necessarily non-periodic) solutions in the standard universe of ZFC (because the tiling equation is aperiodic), but (by the completeness theorem) will not admit solutions at all in some other (nonstandard) universes of ZFC. For the convenience of the reader we review the proof of this assertion (generalized to multiple tiles, and to arbitrary periodic subsets E of explicit ﬁnitely generated abelian groups G) in Appendix A.

1.3. The periodic tiling conjecture. The following conjecture was proposed 3 in the case E = G = Zd by Lagarias and Wang [LW96] and also previously appears implicitly in [GS87, p. 23]: Conjecture 1.5 (Periodic tiling conjecture). Let G be an explicit ﬁnitely generated abelian group, let F be a ﬁnite non-empty subset of G, and let E be a periodic subset of G. Then Tile(F ; E) is not aperiodic.

By the previous discussion, Conjecture 1.5 implies that the tiling equation Tile(F ; E) is decidable for every F,E,G obeying the hypotheses of the conjecture. The following progress is known towards the periodic tiling conjecture: • Conjecture 1.5 is trivial when G is a ﬁnite abelian group, since in this case all subsets of G are periodic. • When E = G = Z, Conjecture 1.5 was established by Newman [N77] as a consequence of the pigeonhole principle. In fact, the argument shows that every set in Tile(F ; Z)U is periodic. As we shall review in Section 2 below, the argument also extends to the case G = Z × G0 for an (explicit) ﬁnite abelian group G0, and to an arbitrary periodic subset E of G. (See also the results in Section 10 for some additional properties of one-dimensional tilings.) • When E = G = Z2, Conjecture 1.5 was established by Bhattacharya 2 [B20] using ergodic theory methods (viewing Tile(F ; Z )U as a dynamical system using the translation action of Z2). In our previous paper [GT20] we gave an alternative proof of this result, and generalized it to the case where E is a periodic subset of G = Z2. In fact, we strengthen the 3Strictly speaking, Lagarias and Wang posed an analogue of this conjecture for E = G = Rd, see Subsection 12.2. UNDECIDABLE TILINGS WITH ONLY TWO TILES 5

previous result of Bhattacharya, by showing that every set in Tile(F,E)U is weakly periodic (the disjoint union of ﬁnitely many one-periodic sets). In the case of polyominoes (where F is viewed as a union of unit squares whose boundary is a simple closed curve), the conjecture was previously established in [BN91], [G-BN91]4 (and decidability was established even earlier in [WvL84]). The conjecture remains open in other cases; for instance, the case E = G = 3 2 Z or the case E = G = Z × ZN for an arbitrary natural number N, are currently unresolved, although we hope to report on some results in these cases in forthcoming work. In [S98] it was shown that Conjecture 1.5 for E = G = Zd was true whenever the cardinality |F | of F was prime, or less than or equal to four.

1.4. Tilings by multiple tiles. It is natural to ask if Conjecture 1.5 extends to tilings by multiple tiles. Given subsets F1,...,FJ ,E of a group G, we use J 5 Tile(F1,...,FJ ; E) = Tile((Fj)j=1; E) to denote the tiling equation

J ] Xj ⊕ Fj = E, (1.3) j=1 where A ] B denotes the disjoint union of A and B (equal to A ∪ B when A, B are disjoint, and undefined otherwise). As before we view F1,...,FJ ,E as given data for this equation, and X1,..., XJ are indeterminate variables representing unknown tiling sets in G. If G is an explicit finitely generated group, F1,...,FJ are finite subsets of G, and E is a periodic subset of G, we can define the solution set ( J ) ] Tile(F1,...,FJ ; E)U := (A1,...,AJ ): A1,...,AJ ⊂ G; Aj ⊕ Fj = E j=1 and more generally for any other universe U∗ of ZFC we have ( J ) ] Tile(F1,...,FJ ; E)U∗ := (A1,...,AJ ): A1,...,AJ ⊂ GU∗ ; Aj ⊕ (Fj)U∗ = EU∗ . j=1

We extend Definition 1.2 to multiple tilings in the natural fashion: Definition 1.6 (Undecidability and aperiodicity for multiple tiles). Let G be an explicit finitely generated abelian group, F1,...,FJ be finite subsets of G for some standard natural number J, and E a periodic subset of G.

(i) We say that the tiling equation Tile(F1,...,FJ ; E) is undecidable if the assertion that there exist subsets A1,...,AJ ⊂ G solving Tile(F1,...,FJ ; E), when phrased as a ﬁrst-order sentence in ZFC, is not provable within the axiom system of ZFC. By the G¨odelcompleteness theorem, this is equivalent to the assertion that Tile(F1,...,FJ ; E)U∗ is non-empty for some

4 2 In fact, they showed that when F is a polyomino, every set in Tile(F ; Z )U is one-periodic. 5See Section 1.8 for our conventions on precedence of operations such as ⊕ and ]. In the 1 1 1 1 language of convolutions, one can also write this tiling equation as X1 ∗ F1 +···+ XJ ∗ FJ = 1E, where we use 1A to denote the indicator function of A. 6 RACHEL GREENFELD AND TERENCE TAO

universes U∗ of ZFC, but empty for some other universes. We say that Tile(F1,...,FJ ; E) is decidable if it is not undecidable. (ii) We say that the tiling equation Tile(F1,...,FJ ; E) is aperiodic if, when working within the standard universe U of ZFC, the equation Tile(F1,...,FJ ; E) admits a solution A1,...,AJ ⊂ G, but there are no solutions for which all of the A1,...,AJ are periodic. That is to say, Tile(F1,...,FJ ; E)U is non-empty but contains no tuples of periodic sets.

As in the single tile case, undecidability implies aperiodicity; see Appendix A. The argument of Newman that resolves the one-dimensional case of Conjecture 1.5 also shows that for (explicit) one-dimensional groups G = Z × G0, every tiling equation Tile(F1,...,FJ ; E) is not aperiodic (and thus also decidable); see Section 2. However, in marked contrast to what Conjecture 1.5 predicts to be the case for single tiles, it is known that a tiling equation Tile(F1,...,FJ ; E) can be aperiodic or even undecidable when J is large enough. In the model case 2 2 E = G = Z , an aperiodic tiling equation Tile(F1,...,FJ ; Z ) was famously constructed6 by Berger [B66] with J = 20426, and an undecidable tiling was also constructed by a modification of the method with an unspecified value of J. A simplified proof of this latter fact was given by Robinson [R71], who also constructed a collection of J = 36 tiles was constructed in which a related completion problem was shown to be undecidable. The value of J for either undecidable examples or aperiodic examples has been steadily lowered over time; see Table 1 for a partial list. We refer the reader to the recent survey [JV20] for more details of these results. To our knowledge, the smallest known 2 value of J for an aperiodic tiling equation Tile(F1,...,FJ ; Z ) is J = 8, by Ammann, Grunbaum, and Shephard [AGS92]. The smallest known value of 2 J for a tiling equation Tile(F1,...,FJ ; Z ) that was explicitly constructed and shown to be undecidable is J = 11, due to Ollinger [O09]. Remark 1.7. As Table 1 demonstrates, many of these constructions were based on a variant of a tile set in Z2 known as a set of Wang tiles, but in [JR21] it was shown that Wang tile constructions cannot create aperiodic (or undecidable) tile sets for any J < 11.

Analogous constructions in higher dimensions were obtained for E = G = Z3 (or more precisely R3) in [D89], [S97], [CK95] and for E = G = Zn (or more n precisely R ), n > 3 in [G-S99b].

1.5. Main results. Our first main result is that one can in fact obtain undecidable (and hence aperiodic) tiling equations for J as small as 2, at the cost of 2 2 enlarging E from Z to Z × E0 for some subset E0 of a (explicit) finite abelian group G0. 2 Theorem 1.8 (Undecidable tiling equation with two tiles in Z × G0). There exists an explicit finite abelian group G0, a subset E0 of G0, and finite non-empty

6Strictly speaking, Berger’s construction was for the closely related domino problem (or Wang tiling problem), but it was shown by Golomb [G70] shortly afterwards that this construction also implies undecidability for the translational tiling problem. Similar considerations apply to several of the other constructions listed in Table 1. UNDECIDABLE TILINGS WITH ONLY TWO TILES 7

J Author Type 20426 Berger [B66] aperiodic [undecidable] (W) 104 Robinson [R67] aperiodic (W) 104 Ollinger [O08] aperiodic [undecidable] (W) 103 Berger [B64] aperiodic (W) 86 Knuth [K68] aperiodic (W) 56 Robinson [R71] aperiodic (W) 52 Robinson [P80] aperiodic (W) 40 Lauchli [W75] aperiodic (W) 36 Robinson [R71] completion-undecidable (W) 32 Robinson [GS87] aperiodic (W) 24 Grunbaum–Shephard [GS87] aperiodic (W) 24 Robinson [GS87] aperiodic (W) 16 Ammann et al. [AGS92] aperiodic (W) 16 Goodman-Strauss [G-S99] aperiodic 14 Kari [K96] aperiodic (W) 13 Culik [C96] aperiodic (W) 11 Jeandel–Rao [JR21] aperiodic (W) 11 Ollinger [O09] undecidable 8 Ammann et al. [AGS92] aperiodic 2∗ Theorems 1.8. 1.9 undecidable 1∗∗ Theorem 11.2 undecidable Table 1. Selected constructions of aperiodic or undecidable tiling equations. This list is primarily adapted from [JR21], and incorporates from that reference some corrections to the values of J in several lines of this table. Constructions labeled (W) arise from a Wang tile construction. The constructions marked “aperiodic [undecidable]” give aperiodic tilings for the specified value of J, and an undecidable tiling for an unspecified value of J. The asterisk for our results in Theorems 1.8, 1.9 denotes the 2 2 fact that we are replacing Z by Z ×E0 for some subset E0 of an explicit finite abelian group G0, or by a periodic subset of some high-dimensional lattice Zd. The double asterisk indicates that the tiling is nonabelian. For some other notable constructions of aperiodic or undecidable tiling equations (but with values of J that are either not explicitly stated, or larger than other contem- porary constructions), see [JR21], [G-S99], [JV20].

2 2 subsets F1,F2 of Z × G0 such that the tiling equation Tile(F1,F2; Z × E0) is undecidable (and hence aperiodic).

The proof of Theorem 1.8 goes on throughout Sections 3–8. In Section 9, by “pulling back” the proof of Theorem 1.8, we prove the following analogue in Zd. Theorem 1.9 (Undecidable tiling equation with two tiles in Zd). There exists an explicit d > 1, a periodic subset E of Zd, and ﬁnite non-empty subsets d F1,F2 of Z such that the tiling equation Tile(F1,F2; E) is undecidable (and hence aperiodic). 8 RACHEL GREENFELD AND TERENCE TAO

Remark 1.10. One can further extend our construction in Theorem 1.9 to the d d Euclidean space R , as follows. First, replace each tile Fj ⊂ Z , j = 1, 2, with ˜ d a ﬁnite union Fj of unit cubes centered in Fj, and similarly replace E ⊂ Z with a periodic set E˜ ⊂ Rd. Next, in order to make the construction rigid in the Euclidean space, add “bumps” on the sides (as in the proof of Lemma 9.3). ˜ ˜ ˜ When one does so, the only tilings of E by the F1, F2 arise from tilings of E by F1,F2, possibly after applying a translation, and hence the undecidability of the former tiling problem is equivalent to that of the latter.

Our construction can in principle give a completely explicit description of the sets G0,E0,F1,F2, but they are quite complicated (and the group G0 is large), and we have not attempted to optimize the size and complexity of these sets in order to keep the argument as conceptual as possible. Remark 1.11. Our argument establishes an encoding for any tiling problem 2 2 Tile(F1,...,FJ ; Z ) with arbitrary number of tiles in Z as a tiling problem 2 with two tiles in Z ×G0. However, in order to prove Theorem 1.8 we only need to be able to encode Wang tilings. Remark 1.12. A slight modification of the proof of Theorem 1.8 also establishes the slightly stronger claim that the decision problem of whether the tiling 2 equation Tile(F1,F2; Z × E0) is solvable for a given finite abelian group G0, 2 given finite non-empty subsets F1,F2 ⊂ Z ×G0 and E0 ⊂ G0, is algorithmically undecidable. Similarly for Theorems 1.9, 11.2 below. This is basically because the original undecidability result of Berger [B66] that we rely on is also phrased in the language of algorithmic undecidability; see Footnote 9 in Section 8. We leave the details of the appropriate modification of the arguments in the context of algorithmic decidability to the interested reader.

Theorem 1.8 supports the belief7 that the tiling problem is considerably less well behaved for J > 2 than it is for J = 1. As another instance of this belief, the J = 1 tilings enjoy a dilation symmetry (see [B20, Proposition 3.1], [GT20, Lemma 3.1], [T95]) that have no known analogue for J > 2. We present a further distinction between the J = 1 and J > 2 situations in Section 10 below, where we show that in one dimension the J = 1 tilings exhibit a certain partial rigidity property that is not present in the J > 2 setting, and makes any attempt to extend our methods of proof of Theorem 1.8 to the J = 1 case diﬃcult. On the other hand, if one allows the group G0 to be nonabelian, then we can reduce the two tiles in Theorem 1.8 to a single tile: see Section 11.

1.6. Overview of proof. The proof of Theorem 1.8 proceeds by a series of reductions in which we successively replace the tiling equation (1.2) by a more tractable system of equations; see Figure 1.1. We first extend Definition 1.6 to systems of tiling equations. Definition 1.13 (Undecidability and aperiodicity for systems of tiling equations with multiple tiles). Let G be an explicit finitely generated abelian group, J, M > 1 be standard natural numbers, and for each m = 1,...,M, let

7As a dissenting view, it was conjectured in [G-BN91] that the translational tiling problem for Z2 with J = 2 is (algorithmically) decidable and not aperiodic. UNDECIDABLE TILINGS WITH ONLY TWO TILES 9

[B66] §8 Thm. 1.20 §7 Thm. 7.1 §7 Thm. 1.19 11.6 §6 Thm. 11.2 Thm. 1.18

§5 §9 Thm. 1.17 Thm. 9.5 4.1 9.4 Thm. 1.16 Thm. 9.2 1.15 9.1 Thm. 1.8 Thm. 1.9

Figure 1.1. The logical dependencies between the undecidability results in this paper (and in [B66]). For each implication, there is listed either the section where the implication is proven, or the number of the key proposition or lemma that facilitates the implication. We also remark that Proposition 9.4 is proven using Lemma 9.3, while Proposition 11.6 is proven using Corollary 11.5, which in turn follows from Lemma 11.4.

(m) (m) (m) F1 ,...,FJ be ﬁnite subsets of G, and let E be a periodic subset of G. (m) (m) (m) (i) We say that the system Tile(F1 ,...,FJ ; E ), m = 1,...,M is undecidable if the assertion that there exist subsets A1,...,AJ ⊂ G that si- (m) (m) (m) multaneously solve Tile(F1 ,...,FJ ; E ) for all m = 1,...,M, when phrased as a ﬁrst-order sentence in ZFC, is not provable within the axiom system of ZFC. That is to say, the solution set M \ (m) (m) (m) Tile(F1 ,...,FJ ; E )U∗ m=1 is non-empty in some universes U∗ of ZFC, and empty in others. We say that the system is decidable if it is not undecidable. (m) (m) (m) (ii) We say that the system Tile(F1 ,...,FJ ; E ), m = 1,...,M is aperiodic if, when working within the standard universe U of ZFC, this system admits a solution A1,...,AJ ⊂ G, but there are no solutions for which all of the A1,...,AJ are periodic. That is to say, the solution set M \ (m) (m) (m) Tile(F1 ,...,FJ ; E )U m=1 10 RACHEL GREENFELD AND TERENCE TAO

is non-empty but contains no tuples of periodic sets. Example 1.14. Let G be an explicit ﬁnitely generated abelian group, and let G0 be a explicit ﬁnite abelian group. The solutions A to the tiling equation Tile({0} × G0; G × G0) are precisely those sets which are graphs A = {(n, f(n)) : n ∈ G} (1.4) for an arbitrary function f : G → G0. It is possible to impose additional con- ditions on f by adding more tiling equations to this “base” tiling equation Tile({0} × G0; G × G0). For instance, if in addition H is a subgroup of G0 and y + H is a coset of H in G0, solutions A to the system of tiling equations

Tile({0} × G0; G × G0), Tile({0} × H; G × (y + H)) are precisely sets A of the form (1.4) where the function f obeys the additional8 constraint f(n) ∈ y + H for all n ∈ G. As a further example, if −y0, y0 are distinct elements of G0, and h is a non-zero element of G, then solutions A to the system of tiling equations

Tile({0} × G0; G × G0), Tile({0, −h} × {0}; G × {−y0, y0}) are precisely sets A of the form (1.4) where the function f takes values in {−y0, y0} and obeys the additional constraint f(n + h) = −f(n) for all n ∈ G. In all three cases one can verify that the system of tiling equations is decidable and not aperiodic.

We then have Theorem 1.15 (Combining multiple tiling equations into a single equation). d Let J, M > 1, let G = Z ×G0 be an explicit finitely generated abelian group for some explicit finite abelian group G0. Let ZN be a cyclic group with N > M, (m) (m) and for each m = 1,...,M let F1 ,...,FJ be finite non-empty subsets of G (m) ˜ ˜ and E0 a subset of G0. Define the finite sets F1,..., FJ ⊂ G × ZN and the ˜ set E0 ⊂ G0 × ZN by M ˜ ] (m) Fj := Fj × {m} (1.5) m=1 and M ˜ ] (m) E0 := E0 × {m}. (1.6) m=1

(m) (m) d (m) (i) The system Tile(F1 ,...,FJ ; Z × E0 ), m = 1,...,M of tiling equa- ˜ ˜ d ˜ tions is aperiodic if and only if the tiling equation Tile(F1,..., FJ ; Z ×E0) is aperiodic. (m) (m) d (m) (ii) The system Tile(F1 ,...,FJ ; Z × E0 ), m = 1,...,M of tiling equa- ˜ ˜ d tions is undecidable if and only if the tiling equation Tile(F1,..., FJ ; Z × ˜ E0) is undecidable.

8In this particular case, the former tiling equation is redundant, being a consequence of the latter. However, we choose to retain the former equation in this example to illustrate the principle of imposing additional constraints on the function f by the insertion of additional tiling equations. UNDECIDABLE TILINGS WITH ONLY TWO TILES 11

Theorem 1.15 can be established by easy elementary considerations; see Sec- tion 3. In view of this theorem, Theorem 1.8 now reduces to the following statement. Theorem 1.16 (An undecidable system of tiling equations with two tiles in 2 Z × G0). There exists an explicit finite abelian group G0, a standard natural number M, and for each m = 1,...,M there exist finite non-empty sets (m) (m) 2 (m) F1 ,F2 ⊂ Z × G0 and E0 ⊂ G0 such that the system of tiling equations (m) (m) 2 (m) Tile(F1 ,F2 ; Z × E0 ), m = 1,...,M is undecidable. The ability to now impose an arbitrary number of tiling equations grants us a substantial amount of flexibility. In Section 4 we will take advantage of this flexibility to replace the system of tiling equations with a system of functional equations, basically by generalizing the constructions provided in Example 1.14. Specifically, we will reduce Theorem 1.16 to the following statement. Theorem 1.17 (An undecidable system of functional equations). There exists an explicit finite abelian group G0, a standard integer M > 1, and for each (m) (m) 2 m = 1,...,M there exist (possibly empty) finite subsets H1 ,H2 of Z × Z2 (m) (m) (m) and (possibly empty sets) F1 ,F2 ,E ⊂ G0 for m = 1,...,M such that 2 the question of whether there exist functions f1, f2 : Z × Z2 → G0 that solve the system of functional equations

] (m) ] (m) (m) (F1 + f1(n + h1)) ] (F2 + f2(n + h2)) = E (1.7) (m) (m) h1∈H1 h2∈H2 2 for all n ∈ Z × Z2 and m = 1,...,M is undecidable (when expressed as a ﬁrst-order sentence in ZFC).

In the above theorem, the functions f1, f2 can range freely in the ﬁnite group G0. By taking advantage of the Z2 factor in the domain, we can restrict f1, f2 D D to range instead in a Hamming cube {−1, 1} ⊂ ZN , which will be more convenient for us to work with, at the cost of introducing an additional sign in the functional equation (1.7). More precisely, in Section 5 we reduce Theorem 1.17 to Theorem 1.18 (An undecidable system of functional equations in the Ham- ming cube). There exists standard integers N > 2 and D,M > 1, and for (m) (m) 2 each m = 1,...,M there exist shifts h1 , h2 ∈ Z and (possibly empty sets) (m) (m) (m) D F1 ,F2 ,E ⊂ ZN for m = 1,...,M such that the question of whether 2 D there exist functions f1, f2 : Z → {−1, 1} that solve the system of functional equations

(m) (m) (m) (m) (m) (F1 + f1(n + h1 )) ] (F2 + f2(n + h2 )) = E (1.8) for all n ∈ Z2, m = 1,...,M, and = ±1 is undecidable (when expressed as a ﬁrst-order sentence in ZFC).

The next step is to replace the functional equations (1.8) with linear equa- 2 tions on Boolean functions fj,d : Z → {−1, 1} (where we now view {−1, 1} as a subset of the integers). More precisely, in Section 6 we reduce Theorem 1.18 to 12 RACHEL GREENFELD AND TERENCE TAO

Theorem 1.19 (An undecidable system of linear equations for Boolean functions). There exists standard integers D > D0 > 1 and M1,M2 > 1, integer (m) coeﬃcients aj,d ∈ Z for j = 1, 2, d = 1,...,D, m = 1,...,Mj, and shifts 2 hd ∈ Z for d = 1,...,D0 such that the question of whether there exist func- 2 tions fj,d : Z → {−1, 1} ⊂ Z for j = 1, 2 and d = 1,...,D that solve the system of linear functional equations D X (m) aj,d fj,d(n) = 0 (1.9) d=1 2 for all n ∈ Z , j = 1, 2, and m = 1,...,Mj, as well as the system of linear functional equations

f2,d(n + hd) = −f1,d(n) (1.10) 2 for all n ∈ Z and d = 1,...,D0, is undecidable (when expressed as a ﬁrst-order sentence in ZFC).

Now that we are working with linear equations for Boolean functions, we can encode a powerful class of constraints, namely all local Boolean constraints. In Section 7 we will reduce Theorem 1.19 to Theorem 1.20 (An undecidable local Boolean constraint). There exist stan- 2 DL dard integers D,L > 1, shifts h1, . . . , hL ∈ Z , and a set Ω ⊂ {−1, 1} 2 such that the question of whether there exist functions fd : Z → {−1, 1}, d = 1,...,D that solve the constraint

(fd(n + hl))d=1,...,D;l=1,...,L ⊂ Ω (1.11) for all n ∈ Z2 is undecidable (when expressed as a ﬁrst-order sentence in ZFC). Finally, in Section 8 we use the previously established existence of undecidable translational tile sets to prove Theorem 1.20, and thus Theorem 1.8.

1.7. Acknowledgments. RG was partially supported by the Eric and Wendy Schmidt Postdoctoral Award. TT was partially supported by NSF grant DMS- 1764034 and by a Simons Investigator Award. We gratefully acknowledge the hospitality and support of the Hausdorﬀ Institute for Mathematics where a signiﬁcant portion of this research was conducted.

1.8. Notation. Given a subset A ⊂ G of an abelian group G and a shift h ∈ G, we deﬁne A + h = h + A := {a + h : a ∈ A}, A − h := {a − h : a ∈ A}, and −A := {−a : a ∈ A}. The unary operator − is understood to take precedence over the binary operator ×, which in turn takes precedence over the binary operator ⊕, which takes precedence over the binary operator ]. Thus for instance A × −B ⊕ −C × D ] E = ((A × (−B)) ⊕ ((−C) × D)) ] E.

By slight abuse of notation, any set of integers will be identiﬁed with the corresponding set of residue classes in a cyclic group ZN , if these classes are distinct. For instance, if M 6 N, we identify {1,...,M} with the residue classes {1 mod N,...,M mod N} ⊂ ZN , and if N > 2, we identify {−1, 1} with {−1 mod N, 1 mod N} ⊂ ZN . UNDECIDABLE TILINGS WITH ONLY TWO TILES 13

2. Periodic tiling conjecture in one dimension

In this section we adapt the pigeonholing argument of Newman [N77] to establish Theorem 2.1 (One-dimensional case of periodic tiling conjecture). Let G = Z × G0 for a some explicit ﬁnite abelian group G0, let J > 1 be a standard integer, let F1,...,FJ be ﬁnite subsets of G, and let E be a periodic subset of G. Then the tiling equation Tile(F1,...,FJ ; E) is not aperiodic (and hence also decidable).

We remark that the same argument also applies to systems of tiling equations in one-dimensional groups Z×G0; this also follows from the above theorem and Theorem 1.15.

Proof. Suppose one has a solution (A1,...,AJ ) ∈ Tile(F1,...,FJ ; E)U to the tiling equation Tile(F1,...,FJ ; E). To establish the theorem it will suﬃce to 0 0 construct a periodic solution A1,...,AJ ∈ Tile(F1,...,FJ ; E)U to the same equation.

We abbreviate the “thickened interval” {n ∈ Z : a 6 n 6 b} × G0 as [[a, b]] for any integers a 6 b. Since the F1,...,FJ are finite, there exists a natural number L such that F1,...,FJ ⊂ [[−L, L]]. Since E is periodic, there exists a natural number r such that E + (n, 0) = E for all n ∈ rZ, where we view (n, 0) as an element of Z × G0. We can assign each n ∈ rZ a “color”, defined as the tuple J ((Aj − (n, 0)) ∩ [[−L, L]])j=1 . This is a tuple of J subsets of the finite set [[−L, L]], and thus there are only finitely many possible colors. By the pigeonhole principle, one can thus find a pair of integers n0, n0 + D ∈ rZ with D > L that have the same color, thus

(Aj − (n0 + D, 0)) ∩ [[−L, L]] = (Aj − (n0, 0)) ∩ [[−L, L]] or equivalently

Aj ∩ [[n0 + D − L, n0 + D + L]] = (Aj ∩ [[n0 − L, n0 + L]]) + (D, 0) (2.1) for j = 1,...,J. 0 We now deﬁne the sets Aj for j = 1,...,J by taking the portion Aj ∩[[n0, n0+ D − 1]] of Aj and extending periodically by DZ × {0}, thus 0 Aj := (Aj ∩ [[n0, n0 + D − 1]]) ⊕ DZ × {0}. Clearly we have the agreement 0 Aj ∩ [[n0, n0 + D − 1]] = Aj ∩ [[n0, n0 + D − 1]] 0 of Aj,Aj on [[n0, n0 + D − 1]], but from (2.1) we also have 0 Aj ∩ [[n0 − L, n0 − 1]] = (Aj ∩ [[n0 + D − L, n0 + D − 1]]) − (D, 0)

= Aj ∩ [[n0 − L, n0 − 1]] and similarly 0 Aj ∩ [[n0 + D, n0 + D + L]] = (Aj ∩ [[n0, n0 + L]]) + (D, 0)

= Aj ∩ [[n0 + D, n0 + D + L]] 14 RACHEL GREENFELD AND TERENCE TAO

0 and thus Aj,Aj in fact agree on a larger region: 0 Aj ∩ [[n0 − L, n0 + D + L]] = Aj ∩ [[n0 − L, n0 + D + L]]. (2.2)

0 0 It will now suﬃce to show that A1, ... , AJ solve the tiling equation Tile(F1,...,FJ ; E). Since both sides of this equation are periodic with respect to translations by DZ × {0}, it suﬃces to establish this claim within [[n0, n0 + D − 1]], that is to say J ] 0 (Aj ⊕ Fj) ∩ [[n0, n0 + D − 1]] = E ∩ [[n0, n0 + D − 1]]. (2.3) j=1

However, since F1,...,FJ are contained in [[−L, L]], so the only portions of 0 0 A1,...,AJ that are relevant for (2.3) are those in [[n0 − L, n0 + D + L − 1]]. 0 But from (2.2) we may replace each Aj in (2.3) by Aj. Since A1,...,AJ solve the tiling equation Tile(F1,...,FJ ; E), the claim follows. Remark 2.2. An inspection of the argument reveals that the hypothesis that G0 was abelian was not used anywhere in the proof, thus Theorem 2.1 is also valid for nonabelian G0 (with suitable extensions to the notation). This gener- alization will be used in Section 11.

3. Combining multiple tiling equations into a single equation

In this section we establish Theorem 1.15. For the rest of the section we use the notation and hypotheses of that theorem. We begin with part (ii). Suppose we have a solution M \ (m) (m) d (m) (A1,...,AJ ) ∈ Tile(F1 ,...,FJ ; Z × E0 )U m=1 (m) (m) d (m) in G to the system of tiling equations Tile(F1 ,...,FJ ; Z × E0 ), m = 1,...,M . If we deﬁne the sets ˜ Aj := Aj × {0} ⊂ G × ZN ˜ ˜ ˜ for j = 1,...,J, then from construction of the Fj we see that A1,..., AJ solve ˜ ˜ d ˜ the single tiling equation Tile(F1,..., FJ ; Z × E0). Conversely, suppose that we have a solution ˜ ˜ ˜ ˜ d ˜ (A1,..., AJ ) ∈ Tile(F1,..., FJ ; Z × E0)U ˜ ˜ d ˜ in G × ZN to the tiling equation Tile(F1,..., FJ ; Z × E0). We claim that ˜ Aj ⊂ G × {0} for all j = 1,...,J. For if this were not the case, then there ˜ would exist j = 1,...,J and an element (g, n) of Aj with n ∈ ZN \{0}. On the (m) ˜ other hand, for any 1 6 m 6 M, the set Fj is non-empty, hence Fj contains an element of the form (fm, m) for some fm ∈ G. By the tiling equation ˜ ˜ d ˜ d ˜ Tile(F1,..., FJ ; Z × E0), we then have (g + fm, n + m) ∈ Z × E0, hence by ˜ construction of E0 we have n + m ∈ {1,...,M} for all m = 1,...,M, or equivalently n + {1,...,M} ⊂ {1,...,M}. UNDECIDABLE TILINGS WITH ONLY TWO TILES 15

But since N > M, this is inconsistent with n being a non-zero element of ZN . ˜ Thus we have Aj ⊂ G × {0} as desired, and we may write ˜ Aj = Aj × {0} ˜ ˜ d for some Aj ⊂ G. By considering the intersection (or “slice”) of Tile(F1,..., FJ ; Z × ˜ E0) with G × {m}, we see that J ] (m) d (m) Aj ⊕ Fj = Z × E0 j=1 for all m = 1,...,M, that is to say A1,...,AJ solves the system of tiling equa- (m) (m) d (m) tions Tile(F1 ,...,FJ ; Z × E0 ), m = 1,...,M. We have thus demon- ˜ ˜ d ˜ strated that the equation Tile(F1,..., FJ ; Z × E0) admits a solution if and (m) (m) d (m) only if the system Tile(F1 ,...,FJ ; Z × E0 ), m = 1,...,M does. This argument is also valid in any other universe U∗ of ZFC, which gives (ii). An in- ˜ ˜ d ˜ spection of the argument also reveals that the equation Tile(F1,..., FJ ; Z ×E0) (m) (m) d admits a periodic solution if and only if the system Tile(F1 ,...,FJ ; Z × (m) E0 ), m = 1,...,M does, which gives (i). As noted in the introduction, in view of Theorem 1.15 we see that to prove Theorem 1.8 it suﬃces to prove Theorem 1.16. This is the objective of the next ﬁve sections of the paper.

Remark 3.1. For future reference we remark that the abelian nature of G0 was not used in the above argument, thus Theorem 1.15 is also valid for nonabelian G0 (with suitable extensions to the notation).

4. From tiling to functions

In this section we reduce Theorem 1.16 to Theorem 1.17, by means of the following general proposition. Proposition 4.1 (Equivalence of tiling equations and functional equations). Let G be an explicit finitely generated abelian group, let G0 be an explicit finite abelian group, let J, M > 1 and N > J be standard natural numbers, and suppose that for each j = 1,...,J and m = 1,...,M one is given a (possibly (m) (m) empty) finite subset Hj of G and a (possibly empty) subset Fj of G0. For (m) each m = 1,...,M, assume also that we are given a subset E of G0. We adopt the abbreviations

[[a]] := {a} × G0; [[a, b]] := {n ∈ Z : a 6 n 6 b} × G0 ⊂ ZN × G0 for integers a 6 b. Then the following are equivalent: (i) The system of tiling equations J (m) (m) (m) Tile −Hj × ({0} × Fj ] [[j]]) ; G × ({0} × E ] [[1,J]]) (4.1) j=1 for all m = 1,...,M, together with the tiling equations J Tile ({0} × [[σ(j)]])j=1 ; G × [[1,J]] (4.2) for every permutation σ : {1,...,J} → {1,...,J} of {1,...,J}, admit a solution. 16 RACHEL GREENFELD AND TERENCE TAO

(ii) There exist fj : G → G0 for j = 1,...,J that obey the system of functional equations

J ] ] (m) (m) (Fj + fj(n + hj)) = E (4.3) j=1 (m) hj ∈Hj for all n ∈ G and m = 1,...,M.

(m) Remark 4.2. The reason why we work with {0} × Fj ] [[j]] instead of just (m) {0} × Fj in (4.1) is in order to ensure that one is working with a non-empty (m) tile (as is required in Theorem 1.16), even when the original tile Fj is empty.

Proof. Let us ﬁrst show that (ii) implies (i). If f1, . . . , fJ obey the system (4.3), we deﬁne the sets A1,...,AJ ⊂ G × ZN × G0 by the formula

Aj = {(n, 0, fj(n)) : n ∈ G}. (4.4)

For any j = 1,...,J and permutation σ : {1,...,J} → {1,...,J}, we have

Aj ⊕ {0} × [[σ(j)]] = G × [[σ(j)]] which gives the tiling equation (4.2) for any permutation σ. Next, for j = 1,...,J and m = 1,...,M, we have

(m) (m) ] ] (m) Aj ⊕−Hj ×({0}×Fj ][[j]]) = {n}× {0}×(Fj +fj(n+hj)][[j]]) n∈G (m) hj ∈Hj (4.5) and the tiling equation (4.1) then follows from (4.3). This shows that (ii) implies (i).

Now assume conversely that (i) holds, thus we have sets A1,...,AJ ⊂ G × ZN × G0 obeying the system of tiling equations (4.1), (4.2). We ﬁrst adapt an argument from Section 3 to claim that each Aj is contained in G × [[0]]. For if this were not the case, there would exist j = 1,...,J and an element (g, n, g0) of Aj with n ∈ ZN \{0}. The left-hand side of the tiling equation (4.2) would then contain (g, n + σ(j), g0), and thus we would have n + σ(j) ∈ {1,...,J} for all permutations σ, thus n + {1,...,J} ⊂ {1,...,J}.

But this is inconsistent with n being a non-zero element of ZN . Thus each Aj is contained in G × [[0]]. If one considers the intersection (or “slice”) of (4.2) with G × [[σ(j)]], we conclude that

Aj ⊕ {0} × [[σ(j)]] = G × [[σ(j)]] for any j = 1,...,J and permutation σ. This implies that for each n ∈ G there is a unique fj(n) ∈ G0 such that (n, 0, fj(n)) ∈ Aj, thus the Aj are of the form (4.4) for some functions fj. The identity (4.5) then holds, and so from (4.1) we obtain the equation (4.3). This shows that (ii) implies (i). UNDECIDABLE TILINGS WITH ONLY TWO TILES 17

The proof of Proposition 4.1 is valid in every universe U∗ of ZFC, thus the solvability question in Proposition 4.1(i) is decidable if and only if the solvability question in Proposition 4.1(ii) is. Applying this fact for J = 2, we see that Theorem 1.17 implies Theorem 1.16. It now remains to establish Theorem 1.17. This is the objective of the next four sections of the paper.

5. Reduction to the Hamming cube

In this section we show how Theorem 1.18 implies Theorem 1.17. Let N, D, (m) (m) (m) (m) (m) M, h1 , h2 , F1 , F2 , E be as in Theorem 1.18. For each d = 1,...,D, D th let πd : ZN → ZN denote the d coordinate projection, thus

y = (π1(y), . . . , πD(y)) (5.1) D for all y ∈ ZN . 2 2 We write elements of Z × Z2 as (n, t) with n ∈ Z and t ∈ Z2. For a pair of ˜ ˜ 2 D functions f1, f2 : Z × Z2 → ZN , consider the system of functional equations −1 ˜ −1 ˜ −1 πd ({0}) + fj(n, t) ] πd ({0}) + fj(n, t + 1) = πd ({−1, 1}) (5.2)

2 for (n, t) ∈ Z × Z2, d = 1,...,D and j = 1, 2, as well as the equations (m) ˜ (m) (m) ˜ (m) (m) F1 + f1((n, t) + (h1 , 0)) ] F2 + f2((n, t) + (h2 , 0)) = E (5.3)

2 for (n, t) ∈ Z × Z2 and m = 1,...,M. It will suﬃce to establish (using an argument formalizable in ZFC) the equivalence of the following two claims: ˜ ˜ 2 D (i) There exist functions f1, f2 : Z × Z2 → ZN solving the system (5.2), (5.3). 2 D (ii) There exist f1, f2 : Z → {−1, 1} solving the system (1.8).

We ﬁrst show that (ii) implies (i). Given solutions f1, f2 to the system (1.8), ˜ ˜ 2 D we deﬁne the functions f1, f2 : Z × Z2 → ZN by the formula ˜ t fj(n, t) := (−1) fj(n) 2 0 1 for j = 1, 2, n ∈ Z , and t ∈ Z2, with the conventions (−1) := 1 and (−1) := −1. The equations (1.8) then imply (5.3), while the fact that the fj takes values in {−1, 1}D implies (5.2) (the key point here is that {−1, 1} = {x} ] {−x} if and only if x ∈ {−1, 1}). This proves that (ii) implies (i). ˜ ˜ 2 D Now we prove (i) implies (ii). Let f1, f2 : Z × Z2 → ZN be solutions to (5.2), (5.3). From (5.2) we see (on applying the projection πd) that ˜ ˜ {πd(fj(n, t))}]{πd(fj(n, t + 1))} = {−1, 1} 2 for all j = 1, 2, d = 1,...,D, and (n, t) ∈ Z × Z2, or equivalently that ˜ πd(fj(n, t)) ∈ {−1, 1} and ˜ ˜ πd(fj(n, t + 1)) = −πd(fj(n, t)) 2 for all j = 1, 2, d = 1,...,D, and (n, t) ∈ Z × Z2. From (5.1), we thus have ˜ D fj(n, t) ∈ {−1, 1} 18 RACHEL GREENFELD AND TERENCE TAO and ˜ ˜ fj(n, t + 1) = −fj(n) 2 for all j = 1, 2 and (n, t) ∈ Z × Z2. Thus we may write ˜ t fj(n, t) = (−1) fj(n) 2 D for some functions fj : Z → {−1, 1} . The system (5.3) is then equivalent to the system of equations (1.8). This shows that (i) implies (ii). These arguments are valid in every universe U∗ of ZFC, thus Theorem 1.18 implies Theorem 1.17. It now remains to establish Theorem 1.18. This is the objective of the next three sections of the paper.

6. Reduction to systems of linear equations on boolean functions

In this section we show how Theorem 1.19 implies Theorem 1.18. Let D, (m) D0, M1, M2, aj,d , hd be as in Theorem 1.19. We let N be a sufficiently large (m) integer. For each j = 1, 2 and m = 1,...,Mj, we consider the subgroup Hj D of ZN defined by ( D ) (m) D X (m) Hj := (y1, . . . , yD) ∈ ZN : aj,d yj = 0 d=1 D and let πd : ZN → ZN for d = 1,...,D be the coordinate projections as in the 2 D D previous section. For some unknown functions f1, f2 : Z → {−1, 1} ⊂ ZN we consider the system of functional equations (m) (m) Hj + fj(n) = Hj (6.1) 2 for all n ∈ Z , j = 1, 2, m = 1,...,Mj, and = ±1, as well as the system −1 −1 −1 (πd ({0}) + f1(n)) ] (πd ({0}) + f2(n + hd)) = πd ({−1, 1}) (6.2) 2 for all n ∈ Z , d = 1,...,D0 and = ±1. It will suffice to establish (using an argument valid in every universe of ZFC) the equivalence of the following two claims:

2 D (i) There exist functions f1, f2 : Z → ZN solving the system (6.1), (6.2). 2 (ii) There exist functions fj,d : Z → {−1, 1} solving the system (1.9), (1.10). 2 We ﬁrst show that (ii) implies (i). Let fj,d : Z → {−1, 1} be solutions to 2 D (1.9), (1.10). We let fj : Z → {−1, 1} be the function

fj(n) := (fj,1(n), . . . , fj,D(n)) (6.3) for n ∈ Z2 and j = 1, 2, where we now view the Hamming cube {−1, 1}D as D lying in ZN . It is then a routine matter to verify that the system of equations (1.9) then implies (6.1), while the system of equations (1.10) implies (6.2). This shows that (ii) implies (i).

Now we show that (i) implies (ii). Let f1, f2 be a solution to the system (6.1), (6.2). We may express fj in components as (6.3), where the fj,d are functions from Z2 to {−1, 1}. From (6.1) we see that (m) (fj,1(n), . . . , fj,D(n)) ∈ Hj UNDECIDABLE TILINGS WITH ONLY TWO TILES 19

d D for all n ∈ Z , j = 1, 2, m = 1,...,Mj (viewing the tuple as an element of ZN ), or equivalently that D X (m) aj,d fj,d(n) = 0 mod N. d=1 PD (m) The left-hand side is an integer that does not exceed d=1 |aj,d | in magnitude, so for N large enough we have D X (m) aj,d fj,d(n) = 0, d=1 that is to say (1.8) holds. Similarly, from (6.2) we see that

{f1,d(n)}]{f2,d(n + hd)} = {−1, 1} 2 for all n ∈ Z and d = 1,...,D0, which gives (1.9). This proves that (i) implies (ii). These arguments are valid in every universe of ZFC, thus Theorem 1.19 implies Theorem 1.18. It now remains to establish Theorem 1.19. This is the objective of the next two sections of the paper.

7. Reduction to a local Boolean constraint

In this section we show how Theorem 1.20 implies Theorem 1.19. (One can also easily establish the converse implication, but we will not need it here.) We begin with some preliminary reductions. We ﬁrst claim that Theorem 1.20 implies a strengthening of itself in which the set Ω can be taken to be symmetric: Ω = −Ω; also, we can take D > 2. To see this, suppose that we can ﬁnd D, L, h1, . . . , hL, Ω obeying the conclusions of Theorem 1.20. We then introduce the symmetric set Ω0 ⊂ {−1, 1}(D+1)L to be the collection of all tuples (ωd,l)d=1,...,D+1;l=1,...,L obeying the constraints

ωD+1,1 = ··· = ωD+1,L and

(ωd,lωD+1,l)d=1,...,D;l=1...,L ∈ Ω. 0 2 Clearly Ω is symmetric. If f1, . . . , fD : Z → {−1, 1} obeys the constraint 2 (1.11), then by setting fD+1 : Z → {−1, 1} to be the constant function fD+1(n) = 1 then we see from construction that 0 (fd(n + hl))d=1,...,D+1;l=1,...,L ∈ Ω (7.1) 2 2 for all n ∈ Z . Conversely, if there was a solution f1, . . . , fD+1 : Z → {−1, 1} to (7.1), then we must have

fD+1(n + h1) = ··· = fD+1(n + hL) and

(fd(n + hl)fD+1(n + hl))d=1,...,D;l=1...,L ∈ Ω, 2 and then the functions fdfD+1 : Z → {−1, 1} for d = 1,...,D would form a solution to (1.11). As these arguments are formalizable in ZFC, we see that Theorem 1.20 for the speciﬁed choice of D, L, h1, . . . , hL, Ω implies Theorem 0 1.20 for D + 1, L, h1, . . . , hL, Ω , giving the claim. 20 RACHEL GREENFELD AND TERENCE TAO

Now consider the system (1.11) for some D, L, h1, . . . , hL, Ω with D > 2 DL 2 and Ω ⊂ {−1, 1} symmetric, and some unknown functions f1, . . . , fD : Z → {−1, 1}. We introduce a variant system involving some other unknown functions fj,d,l with j = 1, 2, d = 1,...,D, l = 1,...,L, consisting of the symmetric Boolean constraint

(f1,d,l(n))d=1,...,D;l=1,...,L ∈ Ω, (7.2) for all n ∈ Z2, the additional symmetric Boolean constraints

f2,d,1(n) = ··· = f2,d,L(n) (7.3) for all n ∈ Z2 and d = 1,...,D, and the shifted constraints

f2,d,l(n + hl) = −f1,d,l(n) (7.4) 2 2 for all n ∈ Z , d = 1,...,D, and l = 1,...,L. Observe that if f1, . . . , fD : Z → 2 {−1, 1} solve (1.11), then the functions fj,d,l : Z → {−1, 1} deﬁned by

f1,d,l(n) := fd(n + hl) and

f2,d,l(n) := −fd(n) 2 obey the system (7.2), (7.3), (7.4); conversely, if fj,d,l : Z → {−1, 1} solve (7.2), (7.3), (7.4), then from (7.3) we have f2,d,l(n) = −fd(n) for all d = 1,...,D, 2 l = 1,...,L, and some functions f1, . . . , fD : Z → {−1, 1}, and then from (7.2), (7.4) we see that f1, . . . , fD solve (1.11). These arguments are formalizable in ZFC, so we conclude that the question of whether the system (7.2), (7.3), (7.4) admits solutions is undecidable. A symmetric set Ω ⊂ {−1, 1}DL can be viewed as the Hamming cube {−1, 1}DL with a ﬁnite number of pairs of antipodal points {, −} removed. The constraint (7.3) is constraining the tuple (f2,d,l(n))d=1,...,D;l=1,...,L to a symmetric subset of {−1, 1}DL, which can thus also be viewed in this fashion. Relabeling fj,d,l as fj,d for d = 1,...,D0 := DL, and assigning the shifts h1, . . . , hL to these labels appropriately, we conclude Theorem 7.1 (An undecidable system of antipode-avoiding constraints). There 2 exist standard integers D0 > 2 and M1,M2 > 1, shifts h1, . . . , hD0 ∈ Z , and (m) D vectors j ∈ {−1, 1} for j = 1, 2 and m = 1,...,Mj such that the question 2 of whether there exist functions fj,d : Z → {−1, 1}, for j = 1, 2, d = 1,...,D0 that solve the constraints (m) (m) (fj,d(n))d=1,...,D0 6∈ {−j , j } (7.5) 2 for all n ∈ Z , j = 1, 2, m = 1,...,Mj, as well as the constraints

f2,d(n + hd) = −f1,d(n) (7.6) 2 for all n ∈ Z and d = 1,...,D0, is undecidable (when expressed as a ﬁrst-order sentence in ZFC).

This is already quite close to Theorem 1.19, except that the linear constraints (1.9) have been replaced by antipode-avoiding constraints (7.5). To conclude the proof of Theorem 1.19, we will show that each antipode-avoiding constraint (7.5) can be encoded as a linear constraint of the form (1.9) after adding some more functions. UNDECIDABLE TILINGS WITH ONLY TWO TILES 21

To simplify the notation we will assume that M1 = M2 = M, which one can (m) assume without loss of generality by repeating the vectors j as necessary. D0 The key observation is the following. If = (1, . . . , D0 ) ∈ {−1, 1} and D0 y1, . . . , yD0 ∈ {−1, 1} , then the following claims are equivalent:

(a)( y1, . . . , yD0 ) 6∈ {−, }.

(b) 1y1 + ··· + D0 yD0 ∈ {−D0 + 2, −D0 + 4,...,D0 − 4,D0 − 2}. 0 0 0 (c) There exist y1, . . . , yD0−2 ∈ {−1, 1} such that 1y1 + ··· + D0 yD0 + y1 + 0 ··· + yD0−2 = 0. Indeed, it is easy to see from the triangle inequality and parity considerations (and the hypothesis D0 > 2) that (a) and (b) are equivalent, and that (b) and (c) are equivalent.

We now set D := D0 +M(D0 −2) and consider the question of whether there 2 exist functions fj,d : Z → {−1, 1}, for j = 1, 2, d = 1,...,D that solve the linear system of equations

D0 D0−2 X (m) X j,d fj,d(n) + fj,D0+(m−1)(D0−2)+d(n) = 0 (7.7) d=1 d=1 for j = 1, 2, m = 1,...,M, n ∈ Z2, as well as the linear system (7.1) for 2 j = 1, 2, n ∈ Z , and d = 1,...,D0. In view of the equivalence of (a) and (c) (and the fact that for each j = 1, 2, m = 1,...,M, and n ∈ Z2, the variables fj,D0+(m−1)(D0−2)+d(n) appear in precisely one constraint, namely the equation (7.7) for the indicated values of j, m, n) we see that this system of equations (7.6), (7.1) admits a solution if and only if the system of equations (7.5), (7.6) admits a solution. This argument is valid in every universe of ZFC, hence the solvability of the system (7.6), (7.1) is undecidable. This completes the derivation of Theorem 1.19 from Theorem 1.20. It now remains to establish Theorem 1.20. This is the objective of the next section of the paper.

8. Undecidability of local Boolean constraints

In this section we prove Theorem 1.20, which by the preceding reductions also establishes Theorem 1.8. Our starting point is the existence of an undecidable tiling equation 2 Tile(F1,...,FJ ; Z ) (8.1) 2 9 for some standard J and some ﬁnite F1,...,FJ ⊂ Z . This was ﬁrst shown in [B66] (after applying the reduction in [G70]), with many subsequent proofs; see for instance [JV20] for a survey. One can for instance take the tile set in [O09], which has J = 11, though the exact value of J will not be of importance here. Note that to any solution 2 (A1,...,AJ ) ∈ Tile(F1,...,FJ ; Z )U

9Berger’s construction is able to encode any instance of the halting problem for Turing machines (with empty input) as a (Wang) tiling problem; since the consistency of ZFC is an undecidable statement equivalent to the non-halting of a certain Turing machine with empty input, this gives the required claim of undecidability. 22 RACHEL GREENFELD AND TERENCE TAO in Z2 of the tiling equation (8.1), one can associate a coloring function c: Z2 → C taking values in the ﬁnite set

J [ C := {j} × Fj j=1 by deﬁning

c(aj + hj) := (j, hj) whenever j = 1,...,J, aj ∈ Aj, and hj ∈ Fj. The tiling equation (8.1) ensures that the coloring function c is well-deﬁned. Furthermore, from construction we see that c obeys the constraint

0 0 c(n) = (j, hj) =⇒ c(n − hj + hj) = (j, hj) (8.2) 2 0 2 for all n ∈ Z , j = 1,...,J, and hj, hj ∈ Fj. Conversely, suppose that c: Z → C is a function obeying (8.2). Then if we deﬁne Aj for each j = 1,...,J to 2 be the set of those aj ∈ Z such that c(aj + hj) = (j, hj) for some hj ∈ Fj, 0 0 from (8.2) we have c(aj + fj) = (j, fj) for all j = 1,...,J, aj ∈ Aj, and 0 fj ∈ Fj, which implies that A1,...,AJ solve the tiling equation (8.1). Thus the solvability of (8.1) is equivalent to the solvability of the equation (8.2); as the former is undecidable in ZFC, the latter is also, since the above arguments are valid in every universe of ZFC. SJ Since the set C = j=1{j} × Fj is ﬁnite, one can establish an explicit bijection ι: C → Ω between this set and some subset Ω of {−1, 1}D for some D. Composing c with this bijection, we see that the question of locating Boolean 2 functions f1, . . . , fD : Z → {−1, 1} obeying the constraints

(f1(n), . . . , fD(n)) ∈ Ω (8.3) and

0 0 0 (f1(n), . . . , fD(n)) = ι(j, hj) =⇒ (f1(n−hj +hj), . . . , fD(n−hj +hj)) = ι(j, hj) (8.4) 2 0 for all n ∈ Z , j = 1,...,J, and hj, hj ∈ Fj, is undecidable in ZFC. However, this set of constraints is of the type considered in Theorem 1.20 (after enumer- 0 0 ating the set of diﬀerences {hj − hj : j = 1,...,J; hj, hj ∈ Fj} as h1, . . . , hL for some L, and combining the various constraints in (8.3), (8.4)), and the claim follows.

9. Proof of Theorem 1.9

In this section we modify the ingredients of the proof of Theorem 1.8 to establish Theorem 1.9. The proofs of both theorems proceed along similar lines, and in fact are both deduced from a common result in Theorem 1.18; see Figure 1.1. We begin by proving the following analogue of Theorem 1.15. Theorem 9.1 (Combining multiple tiling equations into a single equation). Let J, M, d > 1 and N > M be standard natural numbers. For each m = 1,...,M, (m) (m) d (m) let F1 ,...,FJ be ﬁnite non-empty subsets of Z , and let E be a periodic UNDECIDABLE TILINGS WITH ONLY TWO TILES 23

d ˜ ˜ d subset of Z . Deﬁne the ﬁnite sets F1,..., FJ ⊂ Z × {1,...,M} and the periodic set E˜ ⊂ Zd × Z by M ˜ ] (m) Fj := Fj × {m} (9.1) m=1 and M ] (m) E˜ := E × (NZ + {m}). (9.2) m=1

(m) (m) (m) (i) The system Tile(F1 ,...,FJ ; E ), m = 1,...,M of tiling equations ˜ ˜ ˜ is aperiodic if and only if the tiling equation Tile(F1,..., FJ ; E) is aperiodic. (m) (m) (m) (ii) The system Tile(F1 ,...,FJ ; E ), m = 1,...,M of tiling equations ˜ ˜ ˜ is undecidable if and only if the tiling equation Tile(F1,..., FJ ; E) is undecidable.

Proof. We will just prove (i); the proof of (ii) is similar and is left to the reader. The argument will be a “pullback” of the corresponding proof of Theo- (m) (m) (m) rem 1.15(i). First, suppose that the system Tile(F1 ,...,FJ ; E ), m = 1,...,M of tiling equations has a periodic solution A1,...,AJ . Then clearly, the periodic sets ˜ Aj := Aj × NZ, j = 1,...,J ˜ ˜ ˜ constitute a periodic solution for the system Tile(F1,..., FJ ; E). ˜ ˜ ˜ Conversely, suppose that the system Tile(F1,..., FJ ; E) admits a periodic ˜ ˜ ˜ d solution A1,..., AJ . Observe that if Aj ⊂ Z × NZ for each j = 1,...,J, then the “slices”: d ˜ Aj := {a ∈ Z :(a, 0) ∈ Aj}, j = 1,...,J would give a periodic solution to the system of tiling equations (m) (m) (m) Tile(F1 ,...,FJ ; E ), m = 1,...,M.

Now, suppose to the contrary that there is j0 = 1,...,J such that there exists ˜ ˜ ˜ (g, u) ∈ Aj0 where n ∈ Z \ NZ. Observe that since A1,..., AJ is a solution to Tile(F˜ ,..., F˜ ; E˜), we have in particular that for every (f, m) ∈ F˜(m) 6= ∅, 1 J j0 (g + f, u + m) ∈ E.˜ Thus u + m ∈ {1,...,M} ⊕ NZ for every m = 1,...,M. This is only possible ˜ n if u ∈ NZ, a contradiction. Therefore, we have Aj ⊂ Z × NZ, for every j = 1,...,J as needed. As in the proof of Theorem 1.8, Theorem 9.1 allows one to reduce the proof of Theorem 1.9 to proving the following statement. Theorem 9.2 (An undecidable system of tiling equations with two tiles in Zd). There exist standard natural numbers d, M, and for each m = 1,...,M (m) (m) d (m) d there exist ﬁnite non-empty sets F1 ,F2 ⊂ Z and periodic sets E ⊂ Z (m) (m) (m) such that the system of tiling equations Tile(F1 ,F2 ; E ), m = 1,...,M is undecidable. 24 RACHEL GREENFELD AND TERENCE TAO

Figure 9.1. A tiling by the rigid tile R constructed in Lemma 9.3.

We will show that Theorem 1.18 implies Theorem 9.2. In order for the arguments from Section 4 to be effectively pulled back, we will first need to k construct a rigid tile that can encode the finite group Z /Λ0 as the solution set to a tiling equation.

k Lemma 9.3 (A rigid tile). Let N1,...,Nk > 5, and let Λ0 6 Z be the lattice

Λ0 := N1Z × · · · × NkZ. Then there exists a ﬁnite subset R of Zk with the property that the solution set k k Tile(R; Z )U of the tiling equation Tile(R; Z ) consists precisely of the cosets of Λ0, that is to say k k Tile(R; Z )U = Z /Λ0.

Proof. As a ﬁrst guess, one could take R to be the rectangle

R0 := {0,...,N1 − 1} × · · · × {0,...,Nk − 1}. For this choice of R we certainly have that (ii) implies (i). However, (i) will fail to imply (ii) because the tiling Λ0 ⊕ R is not rigid, and it is possible to “slide” portions of this tiling to create additional tilings (cf. Example 1.3). 10 To ﬁx this we need to add and remove some “bumps” to the sides of R0 to prevent sliding. There are many ways to achieve this; we give one such way as follows. For each j = 1, . . . , k, let nj be an integer with 2 6 nj 6 Nj − 3 (the bounds here are in order to keep the “bumps” and “holes” we shall create from touching each other). We form R from R0 by deleting the elements

(n1, . . . , nj−1, 0, nj+1, . . . , nk) from R0 for each j = 1, . . . , k, and then adding the points

(n1, . . . , nj−1,Nj, nj+1, . . . , nk)

10This basic idea of using bumps to create rigidity goes back to Golomb [G70]. UNDECIDABLE TILINGS WITH ONLY TWO TILES 25 back to compensate. Because R was formed from R0 by shifting some elements k of R0 by elements of Λ0, we see that Λ0 ⊕ R = Λ0 ⊕ R0 = Z . Thus we have the inclusion k k Z /Λ0 ⊂ Tile(R; Z )U. k It remains to prove the converse inclusion. Suppose that A ∈ Tile(R; Z )U, thus k k A ⊂ Z and A ⊕ R = Z . Then for any a ∈ A and 1 6 j 6 k, the point

a + (n1, . . . , nj−1, 0, nj+1, . . . , nk) fails to lie in a + R and thus must lie in some other translate a0 + R of R for some a0 ∈ A such that a0 + R is disjoint from a + R. From the construction of R it can be shown after some case analysis (and is also visually obvious, see 0 0 Figure 9.1) that the only possible choice for a is a = a − Njej, where e1, . . . , ek are the standard basis of Zk. Thus the set A is closed under shifts by negative 0 integer linear combinations of N1e1,...,Nkek. If two elements a, a of A lie in diﬀerent cosets of Λ0, then A would contain the set

0 {a, a } ⊕ {−c1N1e1 − · · · − ckNkek : c1, . . . , ck ∈ N} which has density strictly greater than 1 = 1 in the lower left quadrant. N1...Nk |R| This contradicts the tiling equation A ⊕ R = Zk. Thus A must lie in a single k coset y + Λ0 of Λ0. Since we have (y + Λ0) ⊕ R = Z = A ⊕ R, we must then have A = y + Λ0, giving the desired inclusion.

Now we can prove the following analogue of Proposition 4.1.

Proposition 9.4 (Equivalence of tiling equations in G × Zk and functional equations). Let G be an explicitly ﬁnitely generated abelian group, and G0 =

ZN1 × · · · × ZNk be an explicit ﬁnite abelian group with N1,...,Nk > 5. Let J, M > 1 be standard natural numbers, and Λ0,R be as in 9.3. Suppose that for each j = 1,...,J and m = 1,...,M one is given a (possibly empty) ﬁnite subset (m) (m) k Hj of G and a (possibly empty) subset Fj of Z . For each m = 1,...,M, (m) (m) −1 (m) assume also that we are given a subset E0 of G0 and let E := π (E0 ), k where π : Z → G0 is the quotient homomorphism (with kernel Λ0). We adopt the abbreviations

k [[a]] := {a} × R; [[a, b]] := {n ∈ Z : a 6 n 6 b} × R ⊂ Z × Z for integers a 6 b. Let N > J. Then the following are equivalent: (i) The system of tiling equations

J (m) (m) (m) Tile −Hj × ({0} × Fj ] [[j]]) ; G × (NZ × E ] [[1,J]] ⊕ NZ × Λ0) j=1 (9.3) for all m = 1,...,M, together with the tiling equations

J Tile ({0} × [[σ(j)]])j=1 ; G × ([[1,J]] ⊕ NZ × Λ0) (9.4)

for every permutation σ : {1,...,J} → {1,...,J} of {1,...,J}, admit a solution. 26 RACHEL GREENFELD AND TERENCE TAO

(ii) There exist fj : G → G0 for j = 1,...,J that obey the system of functional equations

J ] ] (m) (m) (Fj + fj(n + hj)) = E0 (9.5) j=1 (m) hj ∈Hj for all n ∈ G and m = 1,...,M.

Proof. The proof of the direction (ii) implies (i) is similar to the proof of this direction of Proposition 4.1, with the only diﬀerence that the solution deﬁned there should be pulled back, i.e., one should set

[ −1 k Aj := {n} × NZ × π ({fj(n)}) ⊂ G × NZ × Z n∈G for j = 1,...,J to construct the desired solution to the system (9.3), (9.4). k We turn to prove (i) implies (ii). Let A1,...,AJ ⊂ G × Z × Z be a solution to the systems (9.3), (9.4). As in the proof of Proposition 4.1, by adapting the argument from the proof of Theorem 9.1 once again, one can show that k Aj ⊂ G × NZ × Z . For any n ∈ G and j = 1,...,J, if we then deﬁne the slice k Aj,n ⊂ Z by the formula k Aj,n := {y ∈ Z :(n, 0, y) ∈ Aj} we conclude from (9.4) that

Aj,n ⊕ R = R ⊕ Λ0 which from Lemma 9.3 implies that Aj,n is a coset of Λ0, or equivalently that

−1 Aj,n = π (fj(n))

k for some fj(n) ∈ G0. If one now inspects the G × {0} × Z slice of (4.1), we see that for any m = 1,...,M one has

J ] ] (m) (m) Aj,n+hj ⊕ Fj = E j=1 (m) hj ∈Hj which gives (9.5) upon applying π. This completes the derivation of (ii) from (i).

The proof of Proposition 9.4 is valid in every universe U∗ of ZFC, so in particular the problem in Proposition 9.4(i) is undecidable if and only if the one in Proposition 9.4(ii) is. Hence, to prove Theorem 9.2, it will suffice to 2 establish the following analogue of Theorem 1.17, in which Z × Z2 is pulled back to Z3. Theorem 9.5 (An undecidable system of functional equations in Z2 × Z). There exists an explicit finite abelian group G0, a standard integer M > 1, and (m) (m) for each m = 1,...,M there exist (possibly empty) finite subsets H1 ,H2 of 2 (m) (m) (m) Z × Z and (possibly empty sets) F1 ,F2 ,E ⊂ G0 for m = 1,...,M such UNDECIDABLE TILINGS WITH ONLY TWO TILES 27

2 that the question of whether there exist functions g1, g2 : Z ×Z → G0 that solve the system of functional equations

] (m) ] (m) (m) (F1 + g1(n + h1)) ] (F2 + g2(n + h2)) = E (9.6) (m) (m) h1∈H1 h2∈H2 for all n ∈ Z2 × Z and m = 1,...,M is undecidable (when expressed as a ﬁrst-order sentence in ZFC).

We can now prove this theorem, and hence Theorem 1.9, using Theorem 1.18:

(m) (m) Proof. We repeat the arguments from Section 5. Let N, D, M, h1 , h2 , (m) (m) (m) F1 , F2 , E be as in Theorem 1.18. We recall the systems (5.2), (5.3) of functional equations, introduced in Section 5. D th As before, for each d = 1,...,D, let πd : ZN → ZN denote the d coordinate 2 2 projection. We write elements of Z × Z2 as (n, t) with n ∈ Z and t ∈ Z2 and elements of Z2 × Z as (n, z) with n ∈ Z2 and z ∈ Z. 2 D For a pair of functions g1, g2 : Z ×Z → ZN , consider the system of functional equations

−1 −1 −1 πd ({0}) + gj(n, z) ] πd ({0}) + gj(n, z + 1) = πd ({−1, 1}) (9.7) for d = 1,...,D and j = 1, 2, as well as the equations

(m) (m) (m) (m) (m) F1 + g1((n, z) + (h1 , 0)) ] F2 + g2((n, z) + (h2 , 0)) = E (9.8) for m = 1,...,M. It will suﬃce to establish (using an argument valid in every universe of ZFC) the equivalence of the following two claims:

˜ ˜ 2 D (i) There exist functions f1, f2 : Z × Z2 → ZN solving the systems (5.2), (5.3). 2 D (ii) There exist functions g1, g2 : Z × Z → ZN solving the systems (9.7), (9.8).

Indeed, if (i) is equivalent to (ii), by Section 5, (ii) is equivalent to the 2 D existence of functions f1, f2 : Z → {−1, 1} solving the system (1.8). Hence Theorem 1.18 implies Theorem 9.5. It therefore remains to show that (i) and (ii) are equivalent. Suppose ﬁrst ˜ ˜ 2 D that f1, f2 : Z × Z2 → ZN solve the systems (5.2), (5.3). Then we can deﬁne 2 D g1, g2 : Z × Z → ZN ˜ gj(n, z) = fj(n, z mod 2), j = 1, 2

2 D which solve systems (9.7), (9.8). Conversely, if g1, g2 : Z × Z → ZN solve the ˜ ˜ 2 D systems (9.7), (9.8), then f1, f2 : Z × Z2 → ZN deﬁned by ˜ t fj(n, t) = (−1) gj(n, 0) solve the systems (5.2), (5.3). The claim therefore follows. 28 RACHEL GREENFELD AND TERENCE TAO

10. Single tile versus multiple tiles

In this section we continue the comparison between tiling equations for a single tile J = 1, and for multiple tiles J > 1. In the introduction we have already mentioned the “dilation lemma” [B20, Proposition 3.1], [GT20, Lemma 3.1], [T95] that is a feature of tilings of a single tile F that has no analogue for tilings of multiple tiles F1,...,FJ . Another distinction can be seen by taking the Fourier transform. For simplicity let us consider a tiling equation of the form D Tile(F1,...,FJ ; Z ). In terms of convolutions, this equation can be written as 1 1 1 1 A1 ∗ F1 + ··· + AJ ∗ FJ = 1. Taking distributional Fourier transforms, one obtains (formally, at least) 1 1 1 1 bA1 bF1 + ··· + bAJ bFJ = δ where δ is the Dirac distribution. When J > 1, this equation reveals little 1 about the support properties of the distributions bAj . But when J = 1, the above equation becomes

1bA1bF = δ which now provides signiﬁcant structural information about the Fourier transform of 1A; in particular this Fourier transform is supported in the union of {0} and the zero set of 1bF (which is a trigonometric polynomial). Such information is consistent with the known structural theorems about tiling sets arising from a single tile; see e.g., [GT20, Remark 1.8]. Such a rich structural theory does not seem to be present when J > 2. Now we present a further structural property of tilings of one tile that is not present for tilings of two or more tiles, which we call a “swapping property”. We will only state and prove this property for one-dimensional tilings, but it is conceivable that analogues of this result exist in higher dimensions.

Theorem 10.1 (Swapping property). Let G0 be a ﬁnite abelian group, and for any integers a, b we write

[[a]] := {a} × G0 ⊂ Z × G0 and [[a, b]] := {n ∈ Z : a 6 n 6 b} × G0 ⊂ Z × G0. (0) (1) Let A ,A be subsets of Z × G0 which agree on the left in the sense that A(0) ∩ [[n]] = A(1) ∩ [[n]] whenever n 6 −n0 for some n0. Suppose also that there is a ﬁnite subset F of Z × G0 such that A(0) ⊕ F = A(1) ⊕ F. (10.1) Then we also have A(ω) ⊕ F = A(0) ⊕ F for any function ω : Z → {0, 1}, where [ A(ω) := A(ω(n)) ∩ [[n]] (10.2) n∈Z (0) (1) is a subset of Z × G0 formed by mixing together the ﬁbers of A and A . UNDECIDABLE TILINGS WITH ONLY TWO TILES 29

(j) Proof. For any n ∈ Z and j = 0, 1, we deﬁne the slices An ,Fn ⊂ G by the formulae (j) (j) An := {x ∈ G0 :(n, x) ∈ A } and

Fn := {x ∈ G0 :(n, x) ∈ F }. By inspecting the intersection (or “slice”) of (10.1) at [[n]] for some integer n, we see that ] (0) ] (1) An−l ⊕ Fl = An−l ⊕ Fl. l∈Z l∈Z (Note that all but finitely many of the terms in these disjoint unions are empty.) In terms of convolutions on the finite abelian group G0, this becomes X 1 1 X 1 1 (0) ∗ Fl (x) = (1) ∗ Fl (x) An−l An−l l∈Z l∈Z for all n ∈ Z and x ∈ G0. If one now introduces the functions fn : G0 → C for n ∈ Z by the formula fn := 1 (1) − 1 (0) An An then by hypothesis fn vanishes for n 6 n0, and also X 1 fn−l ∗ Fl (x) = 0 (10.3) l∈Z for every n ∈ Z and x ∈ G. To analyze this equation we perform Fourier analysis on the finite abelian group G0. Let Gb0 be the Pontryagin dual of G0, that is to say the group of homomorphisms ξ : x 7→ ξ · x from G0 to R/Z. For any function f : G0 → C, we define the Fourier transform fb(ξ): Gb0 → C by the formula X fb(ξ) := f(x)e−2πiξ·x.

x∈G0 Applying this Fourier transform to (10.3), we conclude that X 1 fbn−l(ξ)bFl (ξ) = 0 (10.4) l∈Z for all n ∈ Z and ξ ∈ Gb0. 1 Suppose ξ ∈ Gb0 is such that bFl (ξ) is non-zero for at least one integer l. Let lξ be the smallest integer with 1F (ξ) 6= 0, then we can rearrange (10.4) as b lξ

∞ 1F (ξ) X b lξ+l fbn(ξ) = − fdn−l(ξ) 1F (ξ) l=1 b lξ for all integers n. Since fbn(ξ) vanishes for all n 6 n0, we conclude from induction that fbn(ξ) in fact vanishes for all n. 1 To summarize so far, for any ξ ∈ Gb0, either bFl (ξ) vanishes for all l, or else fbn(ξ) vanishes for all n. In either case, we see that we can generalize (10.4) to X 1 ω(n − l)fbn−l(ξ)bFl (ξ) = 0 l∈Z 30 RACHEL GREENFELD AND TERENCE TAO for all n ∈ Z and ξ ∈ Gb0. Inverting the Fourier transform, this is equivalent to X 1 ω(n − l)fn−l ∗ Fl (x) = 0 l∈Z for all n ∈ Z and x ∈ G0, which is in turn equivalent to X 1 1 X 1 1 (0) ∗ Fl (x) = (ω(n−l)) ∗ Fl (x) An−l An−l l∈Z l∈Z and hence ] (0) ] (ω(n−l)) An−l ⊕ Fl = An−l ⊕ Fl l∈Z l∈Z for all n ∈ Z. This gives (10.2) as desired.

Example 10.2. Let G0 = Z2, F = {0} × Z2, and let (j) (j) A := {(n, a (n)) : n ∈ Z} (0) (1) for j = 0, 1, where a , a : Z → Z2 are two functions that agree at negative (0) (1) integers. Then we have A ⊕ F = A ⊕ F = Z × G0. Furthermore, for any ω : Z → {0, 1}, the set (ω) (ω) A := {(n, a (n)) : n ∈ Z} satisﬁes the same tiling equation: (ω) (0) (1) A ⊕ F = A ⊕ F = A ⊕ F = Z × G0.

Example 10.3. Let G0 = Z2, F = {(0, 0), (1, 1)}, and let (j) A := {(n, j): n ∈ Z} for j = 0, 1. Then, as in the previous example, we have A(0) ⊕ F = A(1) ⊕ F = Z × G0. But for any non-constant function ω : Z → {0, 1}, the set (ω) (ω) A := {(n, a (n)) : n ∈ Z} will not obey the same tiling equation: (ω) (0) (1) A ⊕ F 6= A ⊕ F = A ⊕ F = Z × G0. The problem here is that A(0),A(1) do not agree to the left. Thus we see that this hypothesis is necessary for the theorem to hold.

Informally, Theorem 10.1 asserts that if E ⊂ Z × G0 for a finite abelian group G0 and F is a finite subset of Z×G0, then the solution space Tile(F ; E)U to the tiling equation Tile(F ; E) has the following “swapping property”: any two solutions in this space that agree on one side can interchange their fibers arbitrarily and remain in the space. This is quite a strong property that is not shared by many other types of equations. Consider for instance the simple equation

f2(n + 1) = −f1(n) (10.5) constraining two Boolean functions f1, f2 : Z → {−1, 1}; this is a speciﬁc case of the equation (1.10). We observe that this equation does not obey the swapping (0) (0) (1) (1) property. Indeed, consider the two solutions (f1 , f2 ), (f1 , f2 ) to (10.5) given the formula (i) 1n>i+j fj (n) = (−1) UNDECIDABLE TILINGS WITH ONLY TWO TILES 31 for i = 0, 1 and j = 1, 2. These two solutions agree on the left, but for a given function ω : Z → {0, 1}, the swapped functions

(ω) 1n>ω(n)+j fj (n) = (−1) only obeys (10.5) when ω(1) = ω(2). Because of this, unless the equations (10.5) are either trivial or do not admit any two diﬀerent solutions that agree on one side, it does not seem possible to encode individual constraints such as (10.5) inside tiling equations Tile(F ; E) involving a single tile F , at least in one dimension. As such constraints are an important component of our arguments, it does not seem particularly easy to adapt our methods to construct undecidable or aperiodic tiling equations for a single tile. We remark that in the very special case of deterministic tiling equations, such as the aperiodic tiling equations that encode the construction of Kari in [K96], this obstruction is not present, for then if two solutions to (10.5) agree on one side, they must agree everywhere. So it may still be possible to encode such equations inside tiling equations that consist of one tile. However, as was shown in the previous sections, we can encode any system of equations of the type (10.5) in a system of tiling equations involving more than one tile.

Example 10.4. In the group Z × Z4, the solutions to the system of tiling equations

Tile({(0, 0), (0, 2)}, {(0, 1), (0, 3)}; Z × Z4); Tile({(0, 0)}, {(−1, 0)}; Z × {−1, 1}) can be shown to be given precisely by sets A1,A2 ⊂ Z × Z4 of the form

Aj = {(n, fj(n)) : n ∈ Z} for j = 1, 2 and functions f1, f2 : Z → {−1, 1} solving (10.5). The above discussion then provides a counterexample that demonstrates that Theorem 10.1 fails when working with a pair of tiles F1,F2 rather than a single tile.

The obstruction provided by Theorem 10.1 relies crucially on the abelian nature of G0 (in order to utilize the Fourier transform), suggesting that this obstruction is not present in the nonabelian setting. This suggestion is validated by the results in Section 11 below.

11. A nonabelian analogue

In this section we give an analogue of Theorem 1.8 in which we are able to use just one tile instead of two, at the cost of making the group G somewhat nonabelian. The argument will share several features in common with the proof of Theorem 1.8, in particular both arguments will rely on Theorem 1.19 as a source of undecidability (see Figure 1.1). In order to maintain compatibility with the notation of the rest of the paper we will continue to write nonabelian groups G in additive notation G = (G, +). Thus, we caution that in this section the addition operation + (or ⊕) is not necessarily commutative. Example 11.1 (Nonabelian additive notation for permutations). Consider the 2 2 ﬁnite group S 2 ≡ S , the group of permutations α: → on the order 16 Z4 16 Z4 Z4 32 RACHEL GREENFELD AND TERENCE TAO

2 abelian group Z4; this group will play a key role in the constructions of this section. With our additive notation for groups, we have α + β = α ◦ β for α, β ∈ S 2 , with 0 denoting the identity permutation, mα denoting the Z4 composition of m copies of α, and −α denoting the inverse of α.

The notion of a periodic set continues to make sense for subsets of nonabelian groups (note that every finite index subgroup of G contains a finite index normal subgroup), as does the notation of a tiling equation Tile(F ; E). Our main result is Theorem 11.2 (Undecidable nonabelian tiling with one tile). There exists a 2 D group G of the form G = × S 2 × G0 for some standard natural number 0 Z Z4 and explicit finite abelian group G0, a finite non-empty subset F of G, and a D finite non-empty subset E0 of S 2 ×G0, such that the nonabelian tiling equation Z4 2 Tile(F ; Z × E0) is undecidable. We will derive this result from Theorem 1.19 and some additional preparatory results. The main new idea is to encode the Hamming cube {−1, 1}2 as the solution to a system of tiling equations involving only a single tile in S 2 ; this Z4 2 is ultimately in order to be able to access the reflection permutation ρ ∈ Z4, which will play a crucial role in encoding the equation (1.10) using only one tile rather than two. We first need some additional notation, which we summarize in Figure 11.1.

2 Deﬁnition 11.3 (Notation relating to S 2 and ). Z4 Z4

(i) We let ρ ∈ S 2 denote the reflection permutation ρ(y , y ) := (y , y ). Z4 1 2 2 1 2 (ii) We define the regular representation τ : → S 2 by the formula Z4 Z4 τ(h)(x) := x − h 2 for h, x ∈ Z4. 2 (iii) We define the coordinate function π : S 2 → by Z4 Z4 π(α) := α−1(0, 0), and observe that π(α + β) = β−1(π(α)) (11.1)

for α, β ∈ S 2 . In particular we have Z4 π(α + τ(h)) = π(α) + h (11.2) 2 for all α ∈ S 2 and h ∈ . Z4 Z4 2 2 (iv) We view the Hamming cube {−1, 1} as a coset of the subgroup (2Z4) 2 in Z4, where 2Z4 = {0 mod 4, 2 mod 4} is the order two subgroup of Z4. We let B ⊂ S 2 denote the set Z4 B := π−1({−1, 1}2), (11.3) 2 and let K be the order two subgroup of (2Z4) deﬁned by K := {(0, 0), (0, 2)}. UNDECIDABLE TILINGS WITH ONLY TWO TILES 33

B π {−1, 1}2

ρ ρ

2 2 τ π 2 K (2 ) S 2 Z4 Z4 Z4 Z4

ρ Stab({−1, 1})2

Figure 11.1. Maps between various subgroups (or subsets) of 2 S 2 and . Solid arrows denote group homomorphisms; hooked Z4 Z4 arrows denote injections; double-headed arrows denote surjec- tions; and unlabeled hooked arrows denote inclusions.

2 2 (v)A cycle in the permutation group S 2 is a permutation σ : → such Z4 Z4 Z4 2 that there is an enumeration α1, . . . , α16 of Z4 such that σ(αi) = αi+1 for all i = 1,..., 16 (with the periodic convention α17 = α1). Note that any such cycle generates a cyclic subgroup {0, σ, 2σ, . . . , 15σ} of S 2 of order Z4 16. 2 2 (vi) We let Stab({−1, 1} ) ≡ S12 denote the stabilizer group of {−1, 1} , that is to say the subgroup of S 2 consisting of those permutations that act Z4 trivially on the Hamming cube {−1, 1}2.

We can now state our preliminary encoding lemma.

2 Lemma 11.4 (Encoding {−1, 1} as a system of tiling equations in S 2 ). Let Z4 A be a subset of S 2 . Then the following are equivalent: Z4 (i) A is of the form −1 A = π ({y}) = {α ∈ S 2 : π(α) = y} (11.4) Z4 for some y ∈ {−1, 1}2. (ii) A obeys the tiling equation 2 Tile(τ((2Z4) ); B) (11.5) as well as the tiling equations

Tile({φ, σ, 2σ, . . . , 15σ}; S 2 ) (11.6) Z4 2 for every cycle σ ∈ S 2 and every φ ∈ Stab({−1, 1} ). Z4 Proof. It is easy to see from (11.1), (11.2) that (i) implies (ii). Now suppose conversely that (ii) holds. Then A ⊂ B, and moreover for each β ∈ B there 2 exists a unique element of the coset β + τ((2Z4) ) that lies in A. 2 If φ is an arbitrary element of Stab({−1, 1} ) and σ ∈ S 2 is an arbitrary Z4 cycle, we see from two applications of (11.6) that A ⊕ {φ, σ, 2σ, . . . , 15σ} = A ⊕ {0, σ, 2σ, . . . , 15σ} which on cancelling the terms involving σ gives A ⊕ {φ} = A ⊕ {0}. 34 RACHEL GREENFELD AND TERENCE TAO

That is to say, the set A is invariant with respect to the right action of the group Stab({−1, 1}2). If α ∈ A, then α ∈ B, and hence α({−1, 1}2) must contain the origin (0, 0). Let α, α0 ∈ A be such that the images α({−1, 1}2), α0({−1, 1}2) intersect only at the origin. We claim that this implies that π(α) = π(α0). Indeed, sup- 0 2 pose for contradiction that π(α) 6= π(α ). Then the map σ0 : α({−1, 1} ) → α0({−1, 1}2) defined by 0 σ0(α(y)) := α (y) contains no fixed points (the only possible fixed point would be at the origin, but the assumption π(α) 6= π(α0) prohibits this). Since the domain α({−1, 1}2) 0 2 and range α ({−1, 1} ) of this map only intersect at one point, σ0 also contains 2 2 2 no cycles, and thus one can complete σ0 to a cycleσ ˜ : Z4 → Z4 of Z4. By construction, the permutationsσ ˜ + α and α0 agree on {−1, 1}2, thus σ˜ + α = α0 + φ for some φ ∈ Stab({−1, 1}2). Defining σ := (−α) +σ ˜ + α to be the conjugate ofσ ˜ by α, we see that σ is a cycle with α + σ = α0 + φ, but this contradicts the tiling equation (11.6). Thus π(α) = π(α0) as claimed. Now suppose let α, α0 be arbitrary elements of A, dropping the requirement that α({−1, 1}2), α0({−1, 1}2) intersect only at the origin. The cardinality 2 0 2 2 of α({−1, 1} ) ∪ α ({−1, 1} ) is at most seven; since Z4 has order 16, we can 2 2 then certainly find a four-element subset X of Z4 that intersects α({−1, 1} ) ∪ α0({−1, 1}2) only at the origin. We can then find β ∈ B such that β({−1, 1}2) only intersects α({−1, 1}2) ∪ α0({−1, 1}2) at the origin. Since the coset β + 2 00 τ((2Z4) ) intersects A, we conclude that there exists α ∈ A in this coset such that α00({−1, 1}2) = β({−1, 1}2) only intersects α({−1, 1}2) ∪ α0({−1, 1}2) at the origin. By the previous discussion, we have π(α) = π(α00) and π(α0) = π(α00), hence π(α) = π(α0). We conclude that π is constant on A, thus there exists y ∈ {−1, 1}2 such that

A ⊂ {α ∈ S 2 : π(α) = y}. Z4 Observe that the right-hand side has cardinality 15!, while from (11.6) A must have cardinality exactly 16!/16 = 15!. Thus we must have equality here, giving (i) as claimed.

2 We lift this lemma from S 2 to the slightly larger group S 2 × , to make Z4 Z4 Z4 the encoding of {−1, 1}2 more visible:

2 2 Corollary 11.5 (Encoding {−1, 1} as a system of tiling equations in S 2 × ). Z4 Z4 2 Let A be a subset of S 2 × . Then the following are equivalent: Z4 Z4 (i) A is of the form −1 A = π ({y}) × {y} = {(α, y): α ∈ S 2 , π(α) = y} (11.7) Z4 for some y ∈ {−1, 1}2. UNDECIDABLE TILINGS WITH ONLY TWO TILES 35

(ii) A obeys the tiling equation 2 Tile({(τ(h), h): h ∈ (2Z4) }; {(α, π(α)) : α ∈ B}) (11.8) as well as the tiling equations 2 2 Tile({φ, σ, 2σ, . . . , 15σ} × ; S 2 × ) (11.9) Z4 Z4 Z4 2 for every cycle σ ∈ S 2 and every φ ∈ Stab({−1, 1} ). Z4 Proof. The implication of (ii) from (i) is again a routine application of (11.1), (11.2). Now suppose conversely that (ii) holds. From (11.8) we see that A is contained in the set on the right-hand side of (11.8); in particular A is a graph A = {(α, π(α)) : α ∈ A0} 0 0 for some A ⊂ S 2 . Since A satisﬁes the tiling equations (11.8), (11.9), A Z4 satisﬁes the tiling equations 0 2 {(α + τ(h), π(α) + h): α ∈ A , h ∈ (2Z4) } = {(α, π(α)) : α ∈ B} and 0 2 2 (A ⊕ {φ, σ, 2σ, . . . , 15σ}) × = S 2 × . Z4 Z4 Z4 We conclude that A0 must obey the tiling equations (11.5), (11.6). Applying Lemma 11.4, we see that A0 is of the form (11.4) for some y ∈ {−1, 1}2, and we obtain (i) as required. We enumerate the system (11.8), (11.9) as the system of tiling equations

Tile(F`; E`); ` = 1,...,L (11.10) 2 for some explicit collection F ,...,F ,E ,...,E of subsets of S 2 × (indeed 1 L 1 L Z4 Z4 one has L = 1 + 15! × 12!). Thus the sets (11.7) are precisely the solutions to the tiling system (11.10): L \ −1 2 Tile(F`; E`)U = {π ({y}) × {y} : y ∈ {−1, 1} }. (11.11) `=1 Thus we have successfully encoded the Hamming cube {−1, 1}2 as a system of tiling equations, in a manner that allows the reﬂection map ρ ∈ S 2 to interact Z4 with this encoding. We now use the above corollary to encode the solvability question appearing in Theorem 1.19.

Proposition 11.6 (Encoding linear equations). Let D > D0 > 1 and M1,M2 > (m) 1 be natural numbers, and let integer coefficients aj,d ∈ Z for j = 1, 2, d = 2 1,...,D, m = 1,...,Mj, and shifts hd ∈ Z for d = 1,...,D0. Let N be a multiple of 4 that is sufficiently large depending on all previous data. We define the coordinate projections 0 0 2 D 2 π1, . . . , πD :(ZN ) → ZN 00 00 D π , . . . , π :(S 2 ) → S 2 1 D Z4 Z4 000 000 2 π1 , π2 : ZN → ZN 2 2 in the obvious fashion, while also letting Π: ZN → Z4 be the reduction mod 4 2 map, which is a homomorphism with kernel (4ZN ) ; see Figure 11.2. Then the following statements are equivalent: 36 RACHEL GREENFELD AND TERENCE TAO

2 (i) There exist functions fj,d : Z → {−1, 1} ⊂ Z for j = 1, 2 and d = 1,...,D that solve the system of linear functional equations (1.9) for all 2 n ∈ Z , j = 1, 2, and m = 1,...,Mj, as well as the system of linear 2 functional equations (1.10) for all n ∈ Z and d = 1,...,D0. 2 2 D D (ii) There exists a set A ⊂ × × ( ) × (S 2 ) that simultaneously Z Z2 ZN Z4 solves the following systems of nonabelian tiling equations: 1. The tiling equations (m) 2 (m) D Tile({((0, 0), 0)} × H × Cσ; × 2 × H × S 2 ) (11.12) j Z Z j Z4

for all j = 1, 2 and m = 1,...,M , and cycles σ ∈ S 2 , where j Z4 (m) 2 D Hj 6 (ZN ) is the subgroup ( D ) (m) D 2 D X (m) Hj := (y1,d, y2,d)d=1 ∈ (ZN ) : aj,d yj,d = 0 d=1 D and Cσ ⊂ S 2 is the set Z4 00 −1 D−1 Cσ := (π1 ) ({0, σ, . . . , 15σ}) = {0, σ, . . . , 15σ} × S 2 . Z4 2. The tiling equations 000 0 −1 2 000 0 −1 D Tile({(0, 0)}× 2 ×(π ◦π ) ({0})×Cσ; × 2 ×(π ◦π ) ({−1, 1})×S 2 ) Z j d Z Z j d Z4 (11.13) for all d = 1,...,D, j = 1, 2, and cycles σ ∈ S 2 . Z4 3. The tiling equations 0 2 2 D 00 −1 Tile((Td ] Td); Z × Z2 × (ZN ) × (πd ) (B)) (11.14)

for all d = 1,...,D0, where 2 D 00 −1 Td := {((0, 0), 0)} × (ZN ) × (πd ) (τ(K)) 0 2 D 00 −1 Td := {(−hd, 0)} × (ZN ) × (πd ) (ρ + τ(K)). 4. The tiling equations 2 Tile({((0, 0), 0)} × F`,d; Z × Z2 × E`,d) (11.15)

for all d = 1,...,D0 and ` = 1,...,L, where 2 D D 00 0 F`,d := {(y, ζ) ∈ ( ) × S 2 :(π (ζ), Π(π (y))) ∈ F`} ZN Z4 d d 2 D D 00 0 E`,d := {(y, ζ) ∈ ( ) × S 2 :(π (ζ), Π(π (y))) ∈ E`} ZN Z4 d d

and F`,E` are the sets from (11.10).

Proof. Suppose that (i) holds. The sets −1 π ({y}) = {α ∈ S 2 : π(α) = y} Z4 have the same cardinality 15! for all y ∈ {−1, 1}2, so we may arbitrarily enumerate −1 π ({y}) = {αy,1, . . . , αy,15!} 2 2 for each y ∈ {−1, 1} and some distinct permutations αy,k for y ∈ {−1, 1} , 2 k = 1,..., 15!. We then let A denote the set of all elements of Z × Z2 × 2 D D ({−1, 1} ) × (S 2 ) of the form Z4 D D (n, t, (yn,t,d)d=1, (αyn,t,d,k)d=1) UNDECIDABLE TILINGS WITH ONLY TWO TILES 37

(m) Hj Cσ

2D 2 D D {−1, 1} ( ) (S 2 ) ZN Z4

π0 π0 00 d d πd 2 2 Π 2 π {−1, 1} S 2 ZN Z4 Z4

000 000 πj πj

{−1, 1} ZN τ(K) ρ + τ(K) B

Figure 11.2. Some of the sets and maps mentioned in Propo- sition 11.6.

2 for (n, t) ∈ Z × Z2 and k = 1,..., 15!, where t t 2 yn,t,d := ((−1) f1,d(n), (−1) f2,d(n)) ∈ {−1, 1} . (11.16) We now verify the tiling equations (11.12), (11.13), (11.14), (11.15). For any 2 (n, t) ∈ × and any cycle σ ∈ S 2 , we see from Lemma 11.4 that Z Z2 Z4 D D {(αy ,k) : k = 0,..., 15} ⊕ Cσ = S 2 n,t,d d=1 Z4 and thus for any d = 1,...,D, j = 1, 2, σ one has

(m) ] D (m) D A ⊕ {((0, 0), 0)} × H × Cσ = {(n, t)} × ((yn,t,d) + H ) × S 2 j d=1 j Z4 2 (n,t)∈Z ×Z2 and the equations (11.12) then follows from (1.9). In a similar vein, the set 000 0 −1 A ⊕ {(0, 0)} × Z2 × (πj ◦ πd) ({0}) × Cσ for a given d = 1,...,D, j = 1, 2, σ is equal to ! ] ] 000 0 −1 000 D {n} × 2 × (π ◦ π ) (π (yn,t,d)) × S 2 Z j d j Z4 n∈Z2 t∈Z2 and the equation (11.13) then follows from the observation that (11.16) implies the identity 000 000 {πj (yn,0,d)}]{πj (yn,1,d)} = {−1, 1}. Next, the left-hand side 0 A ⊕ (Td ] Td) of (11.14) is equal to

] 2 D 00 −1 {(n, t)} × (ZN ) × (πd ) (An,t,d ⊕ τ(K) ] An+hd,t,d ⊕ (ρ + τ(K))) 2 (n,t)∈Z ×Z2 where A ⊂ S 2 is the set n,t,d Z4

A := {α : k = 1,..., 15!} = {α ∈ S 2 : π(α) = y }. n,t,d yn,t,d,k Z4 n,t,d Observe that from (11.2) we have that

A ⊕ τ(K) = {α ∈ S 2 : π(α) ∈ y + K} n,t,d Z4 n,t,d 38 RACHEL GREENFELD AND TERENCE TAO and similarly from (11.1), (11.2) (and the involutive nature of ρ) that

A ⊕ (ρ + τ(K)) = {α ∈ S 2 : π(α) ∈ ρ(y ) + K}. n+hd,t,d Z4 n+hd,t,d On the other hand, from the equation (1.10) we have 2 (yn,t,d + K) ] (ρ(yn+hd,t,d) + K) = {−1, 1} and the equation (11.14) follows. Finally we verify (11.15). The left-hand side

A ⊕ {((0, 0), 0)} × F`,d of this equation may be written as n 2 2 D D 00 0 o (n, t, y, ζ) ∈ × 2 × ( ) × S 2 :(π (ζ), Π(π (y))) ∈ An,t,d × {Π(yn,t,d)} ⊕ F` . Z Z ZN Z4 d d But from (11.11) (or Corollary 11.5) we have

An,t,d × {Π(yn,t,d)} ⊕ F` = E` and (11.15) follows. This concludes the derivation of (ii) from (i). 2 Now suppose conversely that (ii) holds. For any (n, t) ∈ Z × Z2, let An,t ⊂ 2 D D ( N ) × S 2 be the ﬁber Z Z4 2 D D An,t := {(y, ζ) ∈ ( ) × S 2 :(n, t, y, ζ) ∈ A}. ZN Z4

Then from the tiling equation (11.15) we see that An,t solves the tiling equation 2 Tile(F`,d,E`,d) for all (n, t) ∈ Z × Z2, ` = 1,...,L and d = 1,...,D. This 00 0 implies that the map (y, ζ) 7→ (πd (ζ), Π(πd(y))) is injective on An,t, and that the image 0 00 0 2 A := {(π (ζ), Π(π (y))) : (y, z) ∈ A } ⊂ S 2 × n,t,d d d n,t Z4 Z4 obeys the tiling equations Tile(F`; E`) for all ` = 1,...,L. Applying (11.11) 2 (or Corollary 11.5), we conclude that there exists yn,t,d ∈ {−1, 1} such that 0 A = {α ∈ S 2 : π(α) = y } × {y }. (11.17) n,t,d Z4 n,t,d n,t,d 0 In particular An,t,d has cardinality 15!, hence An,t has cardinality 15! as well. By construction, we see that for any (y, ζ) ∈ An,t, we have 0 Π(πd(y)) = yn,t,d (11.18) and 00 π(πd (ζ)) = yn,t,d for all d = 1,...,D. Next, from (11.13) we have in particular that 2 000 0 −1 D A ⊂ × 2 × (π ◦ π ) ({−1, 1}) × S 2 Z Z j d Z4 and hence 000 0 πj ◦ πd(y) ∈ {−1, 1} 2 whenever (n, t) ∈ Z × Z2,(y, ζ) ∈ An,t, j = 1, 2, and d = 1,...,D. In 0 2 particular, πd(y) ∈ {−1, 1} , which when combined with (11.18) gives 0 πd(y) = yn,t,d 2 2 2 (where by abuse of notation we view {−1, 1} as embedded in both Z4 and ZN ). Thus we have D y = (yn,t,d)d=1 (11.19) UNDECIDABLE TILINGS WITH ONLY TWO TILES 39 whenever (y, ζ) ∈ An,t. From (11.12) we have

2 (m) D A ⊂ × 2 × H × S 2 Z Z j Z4 which when combined with (11.19) implies that

D (m) (yn,t,d)d=1 ∈ Hj 2 for j = 1, 2 and m = 1,...,Mj, and (n, t) ∈ Z × Z2. If we now introduce the 2 boolean functions fj,d : Z → {−1, 1} by the formula

(f1,d(n), f2,d(n)) := yn,0,d (11.20) for n ∈ Z2 and d = 1,...,D, we conclude that D (m) (f1,d(n), f2,d(n))d=1 ∈ Hj or equivalently that D X (m) aj,d fj,d(n) = 0 mod N d=1 2 for all n ∈ Z , j = 1, 2, and m = 1,...,Mj. For N large enough, we may drop the reduction modulo N as the left-hand side is bounded independently of N, thus D X (m) aj,d fj,d(n) = 0 d=1 in the integers. This gives (1.9). Next, from (11.14) we have 2 D 00 −1 An,t ⊕ (ZN ) × (πd ) (τ(K)) 2 D 00 −1 2 D 00 −1 ]An+hd,t ⊕ (ZN ) × (πd ) (ρ + τ(K)) = (ZN ) × (πd ) (B) 2 for any (n, t) ∈ Z × Z2 and d = 1,...,D0. Applying (11.17), this becomes

{α ∈ S 2 : π(α) = y } ⊕ τ(K) Z4 n,t,d ]{α ∈ S 2 : π(α) = y } ⊕ (ρ + τ(K)) = B. Z4 n+hd,t,d Applying (11.1), (11.2), (11.3), this is equivalent to 2 (yn,t,d + K) ] (ρ(yn+hd,t,d) + K) = {−1, 1} . Specializing to t = 0 and using (11.20), we obtain

{f1,d(n)}]{f2,d(n + hd)} = {−1, 1} which is (1.10). This establishes (i).

(m) By Theorem 1.19, there exist choices of D,D0,M1,M2, αj,d , hd such that the problem in Proposition 11.6(i) is undecidable in ZFC. As the proof of this proposition is valid in every universe U∗ of ZFC, we conclude that for N a suﬃciently large (standard) multiple of 4, the problem in Proposition 11.6(ii) is undecidable in ZFC. Thus, we can ﬁnd an undecidable system of nonabelian tiling equations ˜ 2 ˜ ˜ Tile(F`; Z × E`); ` = 1,..., L 40 RACHEL GREENFELD AND TERENCE TAO

2 2 D D for some non-empty subsets F˜ ,..., F˜˜ of × ×( ) ×(S 2 ) and subsets 1 L Z Z2 ZN Z4 2 D D E˜ ,..., E˜ ˜ of × ( ) × (S 2 ) . Applying Theorem 1.15 (and Remark 3.1), 1 L Z2 ZN Z4 we obtain Theorem 11.2 as desired.

12. Open problems and remarks

12.1. Recall that Conjecture 1.5 is open in dimensions d > 2 (see Subsec- tion 1.3 for further discussion and known results). The following question then naturally arises. Problem 12.1. Let G be a non-trivial ﬁnitely generated abelian group. Do there exist any ﬁnite set F ⊂ G and periodic set E ⊂ G such that the tiling equation Tile(F,E) is aperiodic?

We hope to address this problem in a future work.

12.2. Conjecture 1.5 was originally formulated in [LW96] for G = Rd. It is an interesting question to determine the precise relationship between between the Zd and Rd formulations of these conjecture. d Problem 12.2. Let d > 1. What can be said about Conjecture 1.5 for G = R , given that the conjecture holds in Zd? In the one dimensional case, the two formulations are equivalent (see [LW96]). In the two dimensional case the precise relationship between the discrete and continuous formulations of the periodic tiling conjecture is not known. In [Ken92] Kenyon extended the result in [G-BN91] and proved that the periodic tiling conjecture holds for topological discs in R2. In [GT20] we proved that for any ﬁnite F ⊂ Z2 and periodic E ⊂ Z2, all the solutions to the equation Tile(F,E) are weakly periodic. This implies a similar result for some special types of tile F in R2, by using the construction in Remark 1.10. We hope to extend this class of tiles and consider the higher dimensional case of Problem 12.2 in a future work.

12.3. We suggest several possible improvements of our construction.

• It may be possible to modify our argument to allow E0 in Theorem 1.8 to equal G0.

Problem 12.3. Is there any ﬁnite abelian group G0 for which there ex- 2 ist ﬁnite non-empty sets F1,F2 ⊂ Z × G0 such that the tiling equation 2 Tile(F1,F2; Z × G0) is undecidable? 2 • In [G-S99] a construction of two tiles F1,F2 in R is given in which the tiling equation is aperiodic if one is allowed to apply arbitrary isome- tries (not just translations) to the tiles F1,F2; each tile ends up lying in eight translation classes, so in our notation this is actually an aperiodic construction with J = 2 × 8 = 16. Similarly for the “Ammann A2” construction in [AGS92] (with J = 2 × 4 = 8). It may be possible to adapt the construction used to prove Theorem 1.8 so that the tiles F1,F2 are isometric to each other. UNDECIDABLE TILINGS WITH ONLY TWO TILES 41

Problem 12.4. Does our construction provide an example of a ﬁnite 2 abelian group G0, a subset E0 ⊂ G0, and two ﬁnite sets F1,F2 ⊂ Z × G0 2 which are isometric to each other, such that the tiling equation Tile(F1,F2; Z × E0) is undecidable?

• The ﬁnite abelian group G0 in Theorem 1.8 obtained from our construction is quite large. It would be interesting to optimize the size of G0.

Problem 12.5. Find the smallest finite abelian group G0 for which there 2 exist finite non-empty sets F1,F2 ⊂ Z × G0, and E0 ⊂ G0 such that the 2 tiling equation Tile(F1,F2; Z × E0) is undecidable. • It might be possible to reduce the dimension d in Theorem 1.9 by “fold- ing” more efficiently the finite construction of G0 in Theorem 1.8, into a lower dimensional infinite space. Qd Problem 12.6. Let G0 = i=1 ZNi . Suppose that there exist F1,F2 ⊂ 2 2 Z × G0 and E0 ⊂ G0 such that the tiling equation Tile(F1,F2; Z × E0) is undecidable. Does this imply the existence of d0 < 2 + d such that there 0 0 d0 d0 are finite sets F1,F2 ⊂ Z and a periodic set E ⊂ Z for which the tiling 0 0 equation Tile(F1,F2; E) is undecidable? • In Remark 1.12 we discuss the algorithmic undecidable tiling problem which our argument establishes. In this tiling problem, the finite abelian group G0 is one of the inputs. It might be the a slight modification of our construction would imply the existence algorithmic undecidable tiling 2 problem with two tiles in Z × G0, for a fixed finite abelian group G0.

Problem 12.7. Does there exist any ﬁnite abelian group G0 such that 2 the decision problem of whether the tiling equation Tile(F1,F2; Z × E0) 2 is solvable for any given ﬁnite subsets F1,F2 ⊂ Z × G0 and E0 ⊂ G0, is algorithmically undecidable?

Appendix A. Undecidability and aperiodicity

In this section we give a well-known argument of Wang (see [B66, R71]) that undecidability implies aperiodicity. The argument is usually phrased in the language of algorithmic undecidability, but can be adapted without difficulty to the logical notion of undecidability discussed here. Theorem A.1 (Undecidability implies aperiodicity). Let G be an explicit finitely generated abelian group, J, M > 1 be standard natural numbers, and for each (m) (m) (m) m = 1,...,M, let F1 ,...,FJ be finite subsets of G, and let E be a peri- (m) (m) (m) odic subset of G. If the system Tile(F1 ,...,FJ ; E ) for m = 1,...,M is undecidable, then it is aperiodic.

(m) (m) (m) Proof. We will establish the contrapositive: if the system Tile(F1 ,...,FJ ; E ) for m = 1,...,M fails to be aperiodic, then it must be decidable. By deﬁnition of aperiodicity, one of the following two statements must hold:

TM (m) (m) (m) (i) The standard solution set m=1 Tile(F1 ,...,FJ ; E )U is empty. TM (m) (m) (m) (ii) The standard solution set m=1 Tile(F1 ,...,FJ ; E )U contains a periodic tuple (A1,...,AJ ). 42 RACHEL GREENFELD AND TERENCE TAO

In case (ii), since periodic sets are definable, we have a solution to the system (m) (m) (m) ∗ Tile(F1 ,...,FJ ; E ), m = 1,...,M in every universe U of ZFC, and hence by the Gödel completeness theorem the solvability question is decidable (in the positive). Now suppose that we are in case (i). By the compactness 11 theorem in logic , there must therefore exist a finite subset S of Z2 such that (m) (m) (m) the system Tile(F1 ,...,FJ ; E ), m = 1,...,M is not satisfiable in S, in 2 the sense that there does not exist A1,...,AJ ⊂ Z such that (m) (m) (m) ((A1 ⊕ F1 ) ∩ S) ∪ · · · ∪ ((AJ ⊕ FJ ) ∩ S) = E ∩ S for all m = 1,...,M. This latter assertion can be viewed as unsatisfiable boolean sentence involving the finite number of propositions (n ∈ Aj) for j = 1,...,J, m = 1,...,M, and n ∈ S − F (m). The unsatisfiability of this sentence can be proven in ZFC (simply by exhausting a truth table), and it implies the (m) (m) (m) unsolvability of Tile(F1 ,...,FJ ; E ), m = 1,...,M in every universe of ZFC. By the Gödelcompleteness theorem, we thus see that the solvability of this system is decidable (in the negative). The claim follows.

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UCLA Department of Mathematics, Los Angeles, CA 90095-1555. Email address: [email protected]